After taking the first exam, 15 of the students dropped the class.

A) Statistic
B) Parameter

B) Parameter

The height of 2-year-old maple tree is 28.3 ft.

A) Continuous
B) Discrete

A) Continuous

The sample of spheres categorized from softest to hardest.

A) Interval
B) Ratio
C) Nominal
D) Ordinal

D) Ordinal

Identify the sample and population. Also, determine whether the sample is likely to be representative of the population.

An employee at the local ice cream parlor asks three customers if they like chocolate ice cream.

Sample: the 3 selected customers
Population: all customers

Not representative

Is this accurate data?

An airline company advertises that 100% of their flights are on time after checking 5 randomly selected flights and finding that these 5 were on time.

No, The sample was too small.

Convert 2.5 to an equivalent fraction and percent.

A) 2 1/2, 250%
B) 2, 25%
C) 2, 250%
D) 2 1/2, 25%

A) 2 1/2, 250%

Alex and Juana went on a 116-mile canoe trip with their class. On the first day they traveled 29 miles. What percent of the total distance did they canoe?

A) 25%
B) 0.25%
C) 400%
D) 4%

A) 25%

An advertisement for a heating pad says that it can reduce back pain by 200%. What is wrong with this statement?

If a person's back pain was reduced by 100%, it would be completely eliminated, so it is not possible for a person's back pain to be reduced by more than 100%.

A marketing firm does a survey to find out how many people use a product. Of the one hundred people contacted, fifteen said they use the product.

A) Experiment
B) Observational study

B) Observational study

A quality control specialist compares the output from a machine with a new lubricant to the output of machines with the old lubricant.

A) Observational study
B) Experiment

B) Experiment

49, 34, and 48 students are selected from the Sophomore, Junior, and Senior classes with 496, 348, and 481 students respectively.

A) Systematic
B) Convenience
C) Stratified
D) Cluster
E) Random

C) Stratified

A pollster uses a computer to generate 500 random numbers, then interviews the voters corresponding to those numbers.

A) Cluster
B) Random
C) Stratified
D) Convenience
E) Systematic

B) Random

An education researcher randomly selects 48 middle schools and interviews all the teachers at each school.

A) Stratified
B) Cluster
C) Convenience
D) Random
E) Systematic

B) Cluster

An electronics store receives a shipment of eight boxes of calculators. Each box contains ten calculators. A quality control inspector chooses a box by putting eight identical slips of paper numbered 1 to 8 into a hat, mixing thoroughly and then picking a slip at random. He then chooses a calculator at random from the box selected using a similar method with ten slips of paper in a hat. He repeats the process until he obtains a sample of 5 calculators for quality control testing. Does this sampling plan result in a random sample? Simple random sample? Explain.

A) No; yes. The sample is not random because not all calculators have the same chance of being selected. It is a simple random sample because all samples of 5 calculators have the same chance of being selected.

B) Yes; no. The sample is random because all calculators have the same chance of being selected. It is not a simple random sample because some samples are not possible, such as a sample containing 5 calculators from the same box.

C) No; no. The sample is not random because not all calculators have the same chance of being selected. It is not a simple random sample because some samples are not possible, such as a sample containing 5 calculators from the same box.

D) Yes; yes. The sample is random because all calculators have the same chance of being selected. It is a simple random sample because all samples of 5 calculators have the same chance of being selected.

D) Yes; yes. The sample is random because all calculators have the same chance of being selected. It is a simple random sample because all samples of 5 calculators have the same chance of being selected.

The scores on a recent statistics test are given in the frequency distribution below. Construct the corresponding relative frequency distribution.

On a math test, the scores of 24 students were

95 73 77 69 77 77 95 81 77 67 88 73
73 88 77 73 88 77 73 81 73 88 81 69

Construct a frequency distribution. Use 4 classes beginning with a lower class limit of 60.

The frequency table below shows the number of days off in a given year for 30 police detectives.

Construct a histogram. Use the class midpoints for the horizontal scale. Does the result appear to be a normal distribution? Why or why not?

The distribution does not appear to be normal. It is not bell-shaped and it is not symmetric.

Construct the dotplot for the given data.

A store manager counts the number of customers who make a purchase in his store each day. The data are as follows.

10 11 8 14 7 10 10 11 8 7

Use the data to create a stemplot.

The midterm test scores for the seventh-period typing class are listed below.

85 77 93 91 74 65 68 97 88 59 74 83 85 72 63 79

240 casino patrons, were interviewed as they left the casino. 72 of them said they spent most of the time playing the slots. 72 of them said they played blackjack. 36 said they played craps. 12 said roulette. 12 said poker. The rest were not sure what they played the most. Construct a Pareto chart to depict the gaming practices of the group of casino goers. Choose the vertical scale so that the relative frequencies are represented.

The following data give the distribution of the types of houses in a town containing 24,000 houses.

Capes Garrisons Splits
6000 8400 9600

The pie chart shows the percent of the total population of 61,100 of Springfield living in the given types of housing. Round your result to the nearest whole number.

Find the number of people who live in condos.

A) 9165 people
B) 12,220 people
C) 51,935 people
D) 15 people

A) 9165 people

The graph below shows the number of car accidents occurring in one city in each of the years 2001 through 2006. The number of accidents dropped in 2003 after a new speed limit was imposed. Does the graph distort the data? How would you redesign the graph to be less misleading?

The graph distorts the data because the the vertical scale starts at 60 rather than 0, giving the impression of a large difference in the number of accidents, when actually the number of accidents only varies from 90 to 120. To make the graph less misleading, change the vertical scale so that it begins at 0 and increases in increments of 20.

Find the mean for the given sample data.

Andrew asked seven of his friends how many cousins they had. The results are listed below. Find the mean number of cousins.

18 10 7 14 4 3 8

A) 9.1 cousins
B) 10.7 cousins
C) 10.6 cousins
D) 8.6 cousins

A) 9.1 cousins

mean = average

A store manager kept track of the number of newspapers sold each week over a seven-week period. The results are shown below. Find the median number of newspapers sold.

36 30 201 152 278 242 230

A) 230 newspapers
B) 167 newspapers
C) 201 newspapers
D) 152 newspapers

C) 201 newspapers

put in order and find middle value

Find the mode(s) for the given sample data.

77 52 32 52 29 77

A) 52
B) 53.2
C) 77
D) 77, 52

D) 77, 52

both #'s appear twice

Find the midrange for the given sample data.

49 52 52 52 74 67 55 55

A) 61.5
B) 12.5
C) 25
D) 53.5

A) 61.5

Midrange = (high + low) /2

(74 + 52)/2

The weights (in ounces) of 18 cookies are shown. Find the midrange.

0.63 1.28 0.87 0.99 0.81 1.43
1.28 1.20 0.63 1.45 1.37 1.08
1.37 1.45 0.81 1.37 0.99 0.87

A) 1.130 oz
B) 1.040 oz
C) 1.08 oz
D) 1.030 oz

B) 1.040 oz

Midrange = (high + low) /2
(1.45 +.63)/2

Find the mean and median for each of the two samples, then compare the two sets of results.

A comparison is made between summer electric bills of those who have central air and those who have window units.

Central air: mean = $66.20; median = $65
Window unit: mean = $71.60; median = $84

Window units appear to be significantly more expensive.

Use stat list
Run 1-var stats on each to find mean and median

The test scores of 40 students are summarized in the frequency distribution below. Find the mean score.

A) 77.3
B) 69.6
C) 74.5
D) 73.4

A) 77.3

Use stat list
L1 score (high +low)/2
L2 students
Run 1-var stats

A student earned grades of B, B, A, C, and D. Those courses had these corresponding numbers of credit hours: 4, 5, 1, 5, 4. The grading system assigns quality points to letter grades as follows:
A = 4, B = 3, C = 2, D = 1, and F = 0.
Compute the grade point average (GPA)

A) 9.00
B) 1.37
C) 3.46
D) 2.37

D) 2.37

Use stat list
L1 quality points
L2 hours
Run 1-var stats

Jorge has his own business as a painter. The amounts he made in the last five months are shown below. Find the range for the given sample data.

$2446 $2498 $1566 $2041 $1134

A) $880
B) $1364
C) $932
D) $1312

B) $1364

Range = High - Low
$2498 - $1134

Find the variance for the given data.

-7 7 10 -8 4

A) 67.7
B) 67.6
C) 54.2
D) 68.0

A) 67.7

Variance = standard dev ²
Run 1-var stats to get standard deviation then ²

(8.228000972)²

Christine is currently taking college astronomy. The instructor often gives quizzes. On the past seven quizzes, Christine got the following scores:
54 20 36 23 15 40 59

Find the standard deviation

A) 36
B) 8715.6
C) 10,447
D) 17

D) 17

Run 1-var stats to get standard deviation

The heights of a group of professional basketball players are summarized in the frequency distribution below. Find the standard deviation.

A) 3.2 in.
B) 3.3 in.
C) 2.8 in.
D) 2.9 in.

C) 2.8 in.

Use stat list
L1 Height (high +low)/2
L2 frequency
Run 1-var stats

The race speeds for the top eight cars in a 200-mile race are listed below.

188.1 180.4 189.2 188.4 175.6 177.1 181.6 177.4

Use the range rule of thumb to estimate the standard deviation.

A) 6.8
B) 3.4
C) 1.1
D) 7.5

B) 3.4

range rule of thumb = range/4
(189.2 - 175.6)/4

The amount of Jen's monthly phone bill is normally distributed with a mean of $74 and a standard deviation of $8. What percentage of her phone bills are between $50 and $98?

A) 99.99%
B) 99.7%
C) 68%
D) 95%

B) 99.7%

Empirical Rule
68% are within 1 standard deviation
95% are within 2 standard deviation
99.7% are within 3 standard deviation

(98-74)/8 or
(74-50)/8

The ages of the members of a gym have a mean of 47 years and a standard deviation of 10 years. What can you conclude from Chebyshev's theorem about the percentage of gym members aged between 32 and 62?

A) The percentage is approximately 33.3%
B) The percentage is at least 55.6%
C) The percentage is at least 33.3%
D) The percentage is at most 55.6%

B) The percentage is at least 55.6%

The test scores of 32 students are listed below. Construct a boxplot for the data set.

32 37 41 44 46 48 53 55
57 57 59 63 65 66 68 69
70 71 74 74 75 77 78 79
81 82 83 86 89 92 95 99

Run 1-var stats

Construct a modified boxplot for the data. Identify any outliers.

The weights (in ounces) of 27 tomatoes are listed below.

1.7 2.0 2.2 2.2 2.4 2.5 2.5 2.5 2.6
2.6 2.6 2.7 2.7 2.7 2.8 2.8 2.8 2.9
2.9 2.9 3.0 3.0 3.1 3.1 3.3 3.6 4.2

For data which are heavily skewed to the right, P10 is likely to be closer to the median than P90.
True or false?

A) False
B) True

B) True

"You have one chance in ten of winning the race."
Express as a probability value.

A) 0.90
B) 1
C) 0.5
D) 0.10

D) 0.10

A bag contains 6 red marbles, 3 blue marbles, and 5 green marbles. If a marble is randomly selected from the bag, what is the probability that it is blue?

A)1/11
B)1/3
C)3/14
D)1/5

C)3/14

If one of the results is randomly selected, what is the probability that it is a false positive (test indicates the person has the disease when in fact they don't)? What does this probability suggest about the accuracy of the test?

A) 0.405; The probability of this error is high so the test is not very accurate.
B) 0.0967; The probability of this error is high so the test is not very accurate.
C) 0.145; The probability of this error is high so the test is not very accurate.
D) 0.0260; The probability of this error is low so the test is fairly accurate.

B) 0.0967; The probability of this error is high so the test is not very accurate.

26/269

Of 1936 people who came into a blood bank to give blood, 200 people had high blood pressure.
Estimate the probability that the next person who comes in to give blood will have high blood pressure.

A) 0.071
B) 0.103
C) 0.022
D) 0.154

B) 0.103

200/1936

Two white mice mate. The male has both a white and a black fur-color gene. The female has only white fur-color genes. The fur color of the offspring depends on the pairs of fur-color genes that they receive. Assume that neither the white nor the black gene dominates. List the possible outcomes.

A) WW, WW
B) WW, BB
C) WB, BW
D) WW, BW

D) WW, BW

A spinner has equal regions numbered 1 through 15. What is the probability that the spinner will
stop on an even number or a multiple of 3?

A)2/3
B) 12
C)1/3
D)7/9

A)2/3

5 multiples of 3
5 even that are not multiple of 3

10/15

If one of the 1124 people is randomly selected, find the probability that the person is a man or a heavy smoker.

A) 0.514
B) 0.549
C) 0.483
D) 0.516

D) 0.516

total of men + women heavy smokers / total
(545+35)/1124

A 6-sided die is rolled. Find P(3 or 5).

A)1/36
B)1/6
C)1/3
D) 2

C)1/3

A card is drawn from a well-shuffled deck of 52 cards. Find P(drawing a face card or a 4).

A)2/13
B)4/13
C) 16
D)12/13

B)4/13

12 facecards + 4 fours /52
16/52

100 employees of a company are asked how they get to work and whether they work full time or part time. The figure below shows the results. If one of the 100 employees is randomly selected, find the probability that the person drives alone or cycles to work.
1. Public transportation: 7 full time, 10 part time
2. Bicycle: 4 full time, 5 part time
3. Drive alone: 29 full time, 31 part time
4. Carpool: 6 full time, 8 part time

A) 0.60
B) 0.33
C) 0.64
D) 0.69

D) 0.69

(9 cycle + 60 drive)/100
69/100

In one town, 61% of adults have health insurance. What is the probability that 6 adults selected at random from the town all have health insurance? Round to the nearest thousandth if necessary.

A) 3.66
B) 0.098
C) 0.052
D) 0.61

C) 0.052

.61⁶

You are dealt two cards successively (without replacement) from a shuffled deck of 52 playing cards. Find the probability that both cards are black. Express your answer as a simplified fraction.

A)13/51
B)25/102
C)25/51
D)1/2,652

B)25/102

(26/52)*(25/51)

If two different people are randomly selected from the 933 subjects, find the probability that they
are both heavy smokers.

A) 0.0001778
B) 0.002026
C) 0.006462
D) 0.006383

D) 0.006383

(75/933)*(74/932)

In a batch of 8,000 clock radios 6% are defective. A sample of 8 clock radios is randomly selected without replacement from the 8,000 and tested. The entire batch will be rejected if at least one of those tested is defective. What is the probability that the entire batch will be rejected?

A) 0.0600
B) 0.610
C) 0.125
D) 0.390

D) 0.390

binomcdf
8 = trials
.6 = P
0 = x value
find the compliment

1-.609568

In a blood testing procedure, blood samples from 3 people are combined into one mixture. The mixture will only test negative if all the individual samples are negative. If the probability that an individual sample tests positive is 0.1, what is the probability that the mixture will test positive?

A) 0.271
B) 0.729
C) 0.999
D) 0.00100

A) 0.271

compliment ³
then compliment

(1-.01)³ = .729
1-.729

If one of the 87 flights is randomly selected, find the probability that the flight selected arrived on time.

A)43/87
B)76/87
C)11/76
D) None of the above is correct.

B)76/87

Probability of being late is 11/87
probability of being on time is 1-(11/87)

If one of the 1026 subjects is randomly selected, find the probability that the person chosen is a nonsmoker given that it is a woman.

A) 0.437
B) 0.706
C) 0.543
D) 0.379

B) 0.706

nonsmoking women/total women
389/551

The library is to be given 7 books as a gift. The books will be selected from a list of 16 titles. If each book selected must have a different title, how many possible selections are there?

A) 268,435,456
B) 11,440
C) 57,657,600
D) 112

B) 11,440

16 nCr 7

The organizer of a television show must select 5 people to participate in the show. The participants will be selected from a list of 24 people who have written in to the show. If the participants are selected randomly, what is the probability that the 5 youngest people will be selected?

A)1/42,504
B)1/120
C)1/3
D)1/5,100,480

A)1/42,504

1/(24 nCr 5)

How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6, 7 if repetition of digits is not allowed?

A) 343
B) 5
C) 6
D) 210

D) 210

7 nPr 3
order matters

A class has 8 students who are to be assigned seating by lot. What is the probability that the students will be arranged in order from shortest to tallest? (Assume that no two students are the same height.)

A) 0.00019841
B) 0.1000
C) 0.00024802
D) 0.0000248

D) 0.0000248

1/(8 nPr 8)
order matters

12 wrestlers compete in a competition. If each wrestler wrestles one match with each other wrestler, what are the total numbers of matches?

A) 66
B) 156
C) 132
D) 78

A) 66

12 nCr 2
2 = paired wrestling

The cost of a randomly selected orange

A) Discrete
B) Continuous

A) Discrete

The height of a randomly selected student

A) Continuous
B) Discrete

A) Continuous

Determine whether the following is a probability distribution. If not, identify the requirement that is not satisfied.

Not a probability distribution. The sum of the P(x)'s is not 1, since 1.077  1.000.

The number of golf balls ordered by customers of a pro shop has the following probability distribution. Find the mean

A) μ = 8.46
B) μ = 9
C) μ = 9.06
D) μ = 5.79

A) μ = 8.46

Use 1-Var stats

The probabilities that a batch of 4 computers will contain 0, 1, 2, 3, and 4 defective computers are 0.5729, 0.3424, 0.0767, 0.0076, and 0.0003, respectively. Find the mean

A) μ = 2.00
B) μ = 0.42
C) μ = 0.52
D) μ = 1.09

C) μ = 0.52

Use 1-Var stats

In a certain town, 70% of adults have a college degree. The accompanying table describes the probability distribution for the number of adults (among 4 randomly selected adults) who have a college degree. Find the standard deviation for the probability distribution.

A) σ = 2.95
B) σ = 0.84
C) σ = 0.92
D) σ = 1.06

C) σ = 0.92

Use 1-Var stats

Assume that there is a 0.05 probability that a sports playoff series will last four games, a 0.45 probability that it will last five games, a 0.45 probability that it will last six games, and a 0.05 probability that it will last seven games. Is it unusual for a team to win a series in 4 games?

A) Yes
B) No

A) Yes

.05 or above would be unusual

Find the probability of selecting 12 or more girls.

A) 0.006
B) 0.022
C) 0.001
D) 0.007

D) 0.007

.006+.001+.000

Find the probability of selecting exactly 4 girls.

A) 0.061
B) 0.001
C) 0.022
D) 0.122

A) 0.061

Find the probability of selecting 2 or more girls.

A) 0.999
B) 0.994
C) 0.001
D) 0.006

A) 0.999

compliment of (001 + .000)
1-.001

Ten apples, four of which are rotten, are in a refrigerator. Three apples are randomly selected without replacement. Let the random variable x represent the number chosen that are rotten. Construct a table describing the probability distribution, then find the mean and standard deviation for the random variable x.

μ = 1.200
σ = 0.748

P(x):
0 = 6c3/10c3 = 0.167
1 = (6c2*4c1)/10c3 = 0.5
2 = (6c1*4c2)/10c3 = 0.3
3 = 4c3/10c3 = .033

0 = 6good C 3need / 10total C 3trial
1 = (6good C 2need * 4bad C 1need) / 10total C 3trial
2 = (6good C 1need * 4bad C 2need) / 10total C 3trial
3 = 4bad C 3need / 10total C 3trial

Use 1-Var stats

Suppose you pay $2.00 to roll a fair die with the understanding that you will get back $4.00 for rolling a 2 or a 4, nothing otherwise. What is your expected value?

A) -$0.67
B) -$2.00
C) $4.00
D) $2.00

A) -$0.67

L1 = -2 , (4-2)
L2 = 4/6 , 2/6

Run 1-Var stats

The prizes that can be won in a sweepstakes are listed below together with the chances of winning each one: $3800 (1 chance in 8600); $1700 (1 chance in 5400); $700 (1 chance in 4600); $200 (1 chance in 2600). Find the expected value of the amount won for one entry if the cost of entering is 55 cents.

A) $0.44
B) $0.47
C) $0.91
D) $200

A) $0.44

L1 = -.55 , $3800 , $1700 , $700 , $200
L2 = 1 , 1/8600 , 1/5400 , 1/4600 , 1/2600

Run 1-Var stats

Multiple-choice questions on a test each have 4 possible answers, one of which is correct.
Assume that you guess the answers to 5 such questions.

a. Use the multiplication rule to find the probability that the first 2 guesses are wrong and the last 3 guesses are correct. That is, find P(WWCCC), where C denotes a correct answer and W denotes a wrong answer.

b. Make a complete list of the different possible arrangements of 2 wrong answers and 3 correct answers, then find the probability for each entry in the list.

c. Based on the preceding results, what is the probability of getting exactly 3 correct answers when 5 guesses are made?

a. 0.00879

b. WWCCC
WCWCC
WCCWC
WCCCW
CWWCC
CWCWC
CWCCW
CCWWC
CCWCW
CCCWW
Each of the 10 arrangements has probability 0.00879

c. 0.0879

Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial. Round to three decimal places.

n =12, x = 5, p = 0.25

A) 0.027
B) 0.103
C) 0.082
D) 0.091

B) 0.103

use binompdf

An airline estimates that 90% of people booked on their flights actually show up. If the airline books 71 people on a flight for which the maximum number is 69, what is the probability that the number of people who show up will exceed the capacity of the plane?

A) 0.004
B) 0.022
C) 0.005
D) 0.001

C) 0.005

use binomcdf
71 = trials
.9 = P
69 = X

= .9949
Compliment
1-.9949

A car insurance company has determined that 9% of all drivers were involved in a car accident last year. Among the 11 drivers living on one particular street, 3 were involved in a car accident last year. If 11 drivers are randomly selected, what is the probability of getting 3 or more who were involved in a car accident last year?

A) 0.424
B) 0.057
C) 0.070
D) 0.943

C) 0.070

use binomcdf
11 = trials
.09 = P
2 = X

= .9305
Compliment
1-.9305
=.0694?

An archer is able to hit the bull's-eye 50% of the time. If she shoots 8 arrows, what is the probability that she gets exactly 4 bull's-eyes? Assume each shot is independent of the others.

A) 0.219
B) 0.0625
C) 0.273
D) 0.00391

C) 0.273

use binompdf
8 = trials
.5 = P
4 = X

Suppose that 11% of people are left handed. If 5 people are selected at random, what is the probability that exactly 2 of them are left handed?

A) 0.0105
B) 0.171
C) 0.0853
D) 0.0121

C) 0.0853

use binompdf
5 = trials
.11 = P
2 = X

Find the mean, μ,

n = 2164; p = 0.63

A) μ = 1358.0
B) μ = 1367.0
C) μ = 1363.3
D) μ = 1354.8

C) μ = 1363.3

2164 * 0.63

Find the standard deviation, σ.

n = 47; p = 3/5

A) σ = 0.95
B) σ = 3.36
C) σ = 7.48
D) σ = 6.63

B) σ = 3.36

sqrt(47*(3/5)*(2/5))

On a multiple choice test with 9 questions, each question has four possible answers, one of which is correct. For students who guess at all answers, find the mean for the number of correct answers.

A) 4.5
B) 3
C) 2.3
D) 6.8

C) 2.3

9 * 1/4

The probability is 0.6 that a person shopping at a certain store will spend less than $20. For groups of size 24, find the mean number who spend less than $20.

A) 12.0
B) 8.0
C) 9.6
D) 14.4

D) 14.4

24 * .6

On a multiple choice test with 18 questions, each question has four possible answers, one of which is correct. For students who guess at all answers, find the variance for the number of correct answers.

A) 11.4
B) 33.8
C) 3.4
D) 1.8

C) 3.4

18*(3/5)*(2/5)

Consider as unusual any result that differs from the mean by more than 2 standard
deviations. That is, unusual values are either less than μ - 2σ or greater than μ + 2σ.

A survey for brand recognition is done and it is determined that 68% of consumers have heard of Dull Computer Company. A survey of 800 randomly selected consumers is to be conducted. For such groups of 800, would it be unusual to get 595 consumers who recognize the Dull Computer Company name?

A) Yes
B) No

A) Yes

Mean = 800 * .68 = 544
S = sqrt(800 *.68 * .32) = 13.1939

(595 - 544)/13.1939 = 3.9

What is the probability that the random variable has a value greater than 4?

A) 0.450
B) 0.625
C) 0.500
D) 0.375

C) 0.500

(8-4)*.125

What is the probability that the random variable has a value less than 7.4?

A) 1.0500
B) 0.6750
C) 0.9250
D) 0.8000

C) 0.9250

(7.4-0)*.125

What is the probability that the random variable has a value between 0.1 and 6.2?

A) 0.8875
B) 1.0125
C) 0.7625
D) 0.6375

C) 0.7625

(6.2-.1)*.125

Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1.

A) 0.8485
B) 0.8708
C) 0.8907
D) 0.1292

B) 0.8708

use normalcdf
upper 1.13

Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1.

A) 0.9699
B) 0.0301
C) 0.9398
D) 0.0602

C) 0.9398

use normalcdf
lower -1.88
upper 1.88

Find the indicated z score. The graph depicts the standard normal distribution with mean 0 and standard deviation 1.

Shaded area is 0.0901.

A) -1.26
B) -1.45
C) -1.39
D) -1.34

D) -1.34

invNorm

The probability that z lies between -0.55 and 0.55

A) -0.4176
B) -0.9000
C) 0.4176
D) 0.9000

C) 0.4176

use normalcdf
lower -.55
upper .55

P(z > 0.59)

A) 0.2224
B) 0.7224
C) 0.2190
D) 0.2776

D) 0.2776

use normalcdf
lower .59
upper 99

P(-0.73 < z < 2.27)

A) 0.7557
B) 1.54
C) 0.4884
D) 0.2211

A) 0.7557

use normalcdf
lower -.73
upper 2.27

Find the indicated value.
z0.05

A) 1.545
B) 1.755
C) 1.645
D) 1.325

C) 1.645

invNorm

Find the area of the shaded region. The graph depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler test).

A) 0.6293
B) 0.8051
C) 0.4400
D) 0.7486

D) 0.7486

(110-100)/15
=.6666666667

use normalcdf
lower -99
upper .6666666667

Find the IQ score separating the top 14% from the others.

A) 83.7
B) 104.7
C) 99.8
D) 116.2

D) 116.2

1-.14 = .86
invNorm = 1.0803
(1.0803*15)+100 = 116.2

A bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50. Find P60, the score which separates the lower 60% from the top 40%.

A) 212.5
B) 211.3
C) 207.8
D) 187.5

A) 212.5

invNorm .6 = .2533
(.2533*50)+200

The serum cholesterol levels for men in one age group are normally distributed with a mean of 178.1 and a standard deviation of 40.9. All units are in mg/100 mL. Find the two levels that separate the top 9% and the bottom 9%.

A) 123.3 mg/100mL and 232.9 mg/100mL
B) 161.3 mg/100mL and 194.9 mg/100mL
C) 165.0 mg/100mL and 191.19 mg/100mL
D) 106.9 mg/100mL and 249.3 mg/100mL

A) 123.3 mg/100mL and 232.9 mg/100mL

invNorm .09 = 1.340755
(-1.340755*40.9)+178.1 = 123.263

invNorm (1-.09) = 1.340755
(1.340755*40.9)+178.1 = 232.9

The incomes of trainees at a local mill are normally distributed with a mean of $1100 and a standard deviation of $150. What percentage of trainees earn less than $900 a month?

A) 35.31%
B) 40.82%
C) 90.82%
D) 9.18%

D) 9.18%

(900-1100)/150 = -1.33333333
use normalcdf
lower -99
upper -1.33333333

The weekly salaries of teachers in one state are normally distributed with a mean of $490 and a standard deviation of $45. What is the probability that a randomly selected teacher earns more than $525 a week?

A) 0.7823
B) 0.2823
C) 0.2177
D) 0.1003

C) 0.2177

($525-$490)/$45 = .777
use normalcdf
lower .7777
upper 99

The lengths of human pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. What is the probability that a pregnancy lasts at least 300 days?

A) 0.0179
B) 0.9834
C) 0.0166
D) 0.4834

C) 0.0166

(300-268)/15 = 2.1333
use normalcdf
lower 2.1333
upper 99

A poll of 1700 randomly selected students in grades 6 through 8 was conducted and found that 37% enjoy playing sports. Is the 37% result a statistic or a parameter? Explain.

Statistic, because it is calculated from a sample, not a population.

The amount of snowfall falling in a certain mountain range is normally distributed with a mean of 91 inches, and a standard deviation of 10 inches. What is the probability that the mean annual snowfall during 25 randomly picked years will exceed 93.8 inches?

A) 0.5808
B) 0.0808
C) 0.4192
D) 0.0026

B) 0.0808

New S = 10/sqrt(25)= 2
(93.8-91)/2 = 1.4

use normalcdf
lower 1.4
upper 99

A bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50. If 40 different applicants are randomly selected, find the probability that their mean is above 215.

A) 0.3821
B) 0.4713
C) 0.0287
D) 0.1179

C) 0.0287

New S = 50/sqrt(40)= 7.90569415
(215-200)/7.90569415 = 1.897366596

use normalcdf
lower 1.4
upper 99

Assume that women's heights are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. If 90 women are randomly selected, find the probability that they have a mean height between 62.9 inches and 64.0 inches.

A) 0.9318
B) 0.7248
C) 0.1739
D) 0.0424

A) 0.9318

New S = 2.5/sqrt(90)= .263523
(62.9-63.6)/.263523 = -2.656314629
(64-63.6)/.263523 = 1.517894

use normalcdf
lower -2.656314629
upper 1.517894

A final exam in Math 160 has a mean of 73 with standard deviation 7.8. If 24 students are randomly selected, find the probability that the mean of their test scores is greater than 71.

A) 0.8962
B) 0.9012
C) 0.0008
D) 0.5036

A) 0.8962

New S = 7.8/sqrt(24)= 1.592168
(71-73)/1.592168 = -1.2561

use normalcdf
lower -2.656314629
upper 1.517894

State whether or not it is suitable to use the normal distribution as an approximation.

n = 59 and p = 0.7

A) Normal approximation is suitable.
B) Normal approximation is not suitable.

A) Normal approximation is suitable.

59 * 0.7 = 41.3
59 * 0.3 = 17.7
both > 10 = suitable

State whether or not it is suitable to use the normal distribution as an approximation.

n = 45 and p = 0.9

A) Normal approximation is suitable.
B) Normal approximation is not suitable.

B) Normal approximation is not suitable.

45 * .9 = 40.5
45 * .1 = 4.5
4.5 < 10 = not suitable

Estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution.

A certain question on a test is answered correctly by 22% of the respondents. Estimate the probability that among the next 150 responses there will be at most 40 correct answers.

A) 0.8997
B) 0.9306
C) 0.0694
D) 0.1003

B) 0.9306

binomcdf

Estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution.

In one county, the conviction rate for speeding is 85%. Estimate the probability that of the next 100 speeding summonses issued, there will be at least 90 convictions.

A) 0.1038
B) 0.3962
C) 0.0420
D) 0.8962

A) 0.1038

binomcdf
89 = X
1-.9006

Use the normal distribution to approximate the desired probability.

Find the probability that in 200 tosses of a fair die, we will obtain at least 40 fives.

A) 0.0871
B) 0.3871
C) 0.1210
D) 0.2229

C) 0.1210

binomcdf
39 = X
1-.8777

Use the normal distribution to approximate the desired probability.

Merta reports that 74% of its trains are on time. A check of 60 randomly selected trains shows that 38 of them arrived on time. Find the probability that among the 60 trains, 38 or fewer arrive on time. Based on the result, does it seem plausible that the "on-time" rate of 74% could be correct?

A) 0.0409, yes
B) 0.0409, no
C) 0.0316, yes
D) 0.0316, no

B) 0.0409, no

binomcdf
60 = trials
.74 = P
38 = X

Find the critical value zα/2 that corresponds to a 99% confidence level.

A) 1.645
B) 2.33
C) 2.575
D) 1.96

C) 2.575

(1-.99)/2 = .005
invNorm

The following confidence interval is obtained for a population proportion, p: 0.843 < p < 0.875. Use these confidence interval limits to find the margin of error, E.

A) 0.032
B) 0.859
C) 0.016
D) 0.017

C) 0.016

(.875 - .843)/2

Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level.

95% confidence; n = 250, x = 130

A) 0.0650
B) 0.0743
C) 0.0557
D) 0.0619

D) 0.0619

p = 130/250 = .52
q = 1 - .52 = .48

E = 1.96 * sqrt(.52*.48)/250

Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level.

95% confidence; the sample size is 5700, of which 20% are successes

A) 0.00780
B) 0.0104
C) 0.0137
D) 0.0120

B) 0.0104

x = 5700 * .2 = 1140
q = 1 - .20 = .80

E = 1.96 * sqrt(.20*.80)/5700

Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level.

In a clinical test with 1400 subjects, 420 showed improvement from the treatment. Find the margin of error for the 99% confidence interval used to estimate the population proportion.

A) 0.0180
B) 0.0276
C) 0.0315
D) 0.0240

C) 0.0315

p = 1400 * 420 = .3
q = 1 - .3 = .7

E = 2.576 * sqrt(.3*.7)/1400

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.

n = 128, x = 61; 90% confidence

A) 0.407 < p < 0.547
B) 0.408 < p < 0.546
C) 0.403 < p < 0.551
D) 0.404 < p < 0.550

D) 0.404 < p < 0.550

1-PropZInt

find the minimum sample size required

Margin of error: 0.008; confidence level: 99%; phat and qhat unknown

A) 25,894
B) 15,900
C) 25,901
D) 26,024

C) 25,901

(2.576²)*.25 = 1.658944
1.658944/(.008²)

find the minimum sample size required

Margin of error: 0.04; confidence level: 99%; from a prior study, phat is estimated by 0.07.

A) 156
B) 270
C) 324
D) 11

B) 270

(2.576²)*(.07*.93) = .431989
.431989/(.04²)

A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature.
Construct the 95% confidence interval for the true proportion of all voters in the state who favor
approval.

A) 0.435 < p < 0.508
B) 0.444 < p < 0.500
C) 0.438 < p < 0.505
D) 0.471 < p < 0.472

C) 0.438 < p < 0.505

1-PropZInt

Of 123 adults selected randomly from one town, 26 of them smoke.
Construct a 99% confidence interval for the true percentage of all adults in the town that smoke.

A) 15.1% < p < 27.2%
B) 12.6% < p < 29.7%
C) 13.9% < p < 28.4%
D) 11.7% < p < 30.6%

D) 11.7% < p < 30.6%

1-PropZInt

In a certain population, body weights are normally distributed with a mean of 152 pounds and a standard deviation of 26 pounds. How many people must be surveyed if we want to estimate the percentage who weigh more than 180 pounds? Assume that we want 96% confidence that the error is no more than 3 percentage points.

A) 411
B) 1168
C) 564
D) 890

B) 1168

1-.96/2 (invNorm) = 2.054
(2.054²*.25)/.03²

Find the critical value zα/2 that corresponds to a 98% confidence level.

A) 1.75
B) 2.05
C) 2.575
D) 2.33

D) 2.33

(1->98)/2 = .01
invNorm

find the margin of error E.
Weights of eggs: 95% confidence; n = 45, x = 1.50 oz, σ = 0.20 oz

A) 0.06 oz
B) 0.44 oz
C) 0.05 oz
D) 0.01 oz

A) 0.06 oz

1.96 * (.2/sqrt45) =.058

find a confidence interval for estimating the population μ.

Test scores: n = 76, x = 46.1, σ = 5.7; 98% confidence

A) 44.4 < μ < 47.8
B) 45.0 < μ < 47.2
C) 44.8 < μ < 47.4
D) 44.6 < μ < 47.6

D) 44.6 < μ < 47.6

ZInterval

48 packages are randomly selected from packages received by a parcel service. The sample has a mean weight of 10.1 pounds and a standard deviation of 2.9 pounds. What is the 95% confidence interval for the true mean weight, μ, of all packages received by the parcel service?

A) 9.1 lb < μ < 11.1 lb
B) 9.0 lb < μ < 11.2 lb
C) 9.3 lb < μ < 10.9 lb
D) 9.4 lb < μ < 10.8 lb

C) 9.3 lb < μ < 10.9 lb

TInterval

How many women must be randomly selected to estimate the mean weight of women in one age group. We want 90% confidence that the sample mean is within 2.7 lb of the population mean, and the population standard deviation is known to be 22 lb.

A) 256
B) 180
C) 181
D) 178

B) 180

1.645 * (22/2.7) = 13.403
13.403² = 179.6

find zα/2 or tα/2

98%; n = 7; σ = 27; population appears to be normally distributed.

A) tα/2 = 2.575
B) tα/2 = 1.96
C) zα/2 = 2.05
D) zα/2 = 2.33

D) zα/2 = 2.33

σ is known so Z
(1-.98)/2 = .01
invNorm

95%; n = 11; σ is known; population appears to be very skewed.

A) zα/2 = 1.812
B) tα/2 = 2.228
C) zα/2 = 1.96
D) Neither the normal nor the t distribution applies.

D) Neither the normal nor the t distribution applies.

both must have approximate bell curve - since population appears to be skewed neither applies

find the margin of error.

95% confidence, n = 91, xbar = 24, s = 14.7

A) 3.06
B) 5.26
C) 2.62
D) 2.75

A) 3.06

1.98 * (14.7/sqrt91)

A savings and loan association needs information concerning the checking account balances of its local customers. A random sample of 14 accounts was checked and yielded a mean balance of $664.14 and a standard deviation of $297.29. Find a 98% confidence interval for the true mean checking account balance for local customers.

A) $493.71 < μ < $834.57
B) $455.65 < μ < $872.63
C) $453.59 < μ < $874.69
D) $492.52 < μ < $835.76

C) $453.59 < μ < $874.69

TInterval

The football coach randomly selected ten players and timed how long each player took to perform a certain drill. Determine a 95% confidence interval for the mean time for all players.

7.5 10.3 9.3 8.1 11.1
7.9 6.9 11.4 10.7 12.2

A) 10.80 min < μ < 8.28 min
B) 10.90 min < μ < 8.18 min
C) 8.28 min < μ < 10.80 min
D) 8.18 min < μ < 10.90 min

C) 8.28 min < μ < 10.80 min

Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim.

A manufacturer considers his production process to be out of control when defects exceed 3%. In a random sample of 85 items, the defect rate is 5.9% but the manager claims that this is only a sample fluctuation and production is not really out of control. At the 0.01 level of significance, test the manager's claim.

H0: p = .03
H1: p > .03
Test statistic: z = 1.57
P-value: p = 0.0582
Critical value: z = 2.33
Fail to reject null hypothesis.
There is not sufficient evidence to warrant rejection of the manager's claim that production is not really out of control.

Use 1 PropZTest
invNorm for critical value
P value is greater than significance so fail to reject

Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim.

An article in a journal reports that 34% of American fathers take no responsibility for child care. A researcher claims that the figure is higher for fathers in the town of Littleton. A random sample of 234 fathers from Littleton yielded 96 who did not help with child care. Test the researcher's claim at the 0.05 significance level.

H0: p = 0.34
H1: p > 0.34
Test statistic: z = 2.27
P-value: p = 0.0116
Critical value: z = 1.645
Reject null hypothesis.
There is sufficient evidence to support the researcher's claim that the proportion for fathers in Littleton is higher than 34%.

Use 1 PropZTest
invNorm for critical value

Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim.

In a clinical study of an allergy drug, 108 of the 202 subjects reported experiencing significant relief from their symptoms. At the 0.01 significance level, test the claim that more than half of all those using the drug experience relief.

H0: p = 0.5
H1: p > 0.5
Test statistic: z = 0.99
P-value: p = 0.1611
Critical value: z = 2.33
Fail to reject null hypothesis.
There is not sufficient evidence to support the claim that more than half of all those using the drug experience relief.

Use 1 PropZTest
invNorm for critical value

Find the P-value for the indicated hypothesis test.

In a sample of 88 children selected randomly from one town, it is found that 8 of them suffer from asthma. Find the P-value for a test of the claim that the proportion of all children in the town who suffer from asthma is equal to 11%.

A) -0.2843
B) 0.5686
C) 0.2843
D) 0.2157

B) 0.5686

Use 1 PropZTest

Find the P-value for the indicated hypothesis test.

An airline claims that the no-show rate for passengers booked on its flights is less than 6%. Of 380 randomly selected reservations, 18 were no-shows. Find the P-value for a test of the airline's claim.

A) 0.0746
B) 0.1230
C) 0.3508
D) 0.1492

D) 0.1492

Use 1 PropZTest

Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim.

The health of employees is monitored by periodically weighing them in. A sample of 54 employees has a mean weight of 183.9 lb. Assuming that σ is known to be 121.2 lb, use a 0.10 significance level to test the claim that the population mean of all such employees weights is less than 200 lb.

H0: μ = 200
H1: μ < 200
Test statistic: z = -0.98
P-value: 0.1635
Fail to reject H0.
There is not sufficient evidence to support the claim that the mean is less than 200 pounds.

Use Z-Test

Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim.

A random sample of 100 pumpkins is obtained and the mean circumference is found to be 40.5 cm. Assuming that the population standard deviation is known to be 1.6 cm, use a 0.05 significance level to test the claim that the mean circumference of all pumpkins is equal to 39.9 cm.

H0: μ = 39.9
H1: μ  39.9
Test statistic: z = 3.75
P-value: 0.0002
Reject H0.
There is sufficient evidence to warrant rejection of the claim that the mean equals 39.9 cm.

Use Z-Test

Identify the null hypothesis, alternative hypothesis, test statistic, critical value or P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim.

The mean resting pulse rate for men is 72 beats per minute. A simple random sample of men who regularly work out at Mitch's Gym is obtained and their resting pulse rates (in beats per minute) are listed below. Use a 0.05 significance level to test the claim that these sample pulse rates come from a population with a mean less than 72 beats per minute.
Assume that the standard deviation of the resting pulse rates of all men who work out at Mitch's Gym is known to be 6.6 beats per minute. Use the traditional method of testing hypotheses.

54 61 69 84 74 64 69
70 66 80 59 71 76 63

H0: μ = 72 beats per minute
H1: μ < 72 beats per minute
Test statistic: z = -1.94
P-value: 0.026
Critical-value: z = -1.645
Reject H0
At the 5% significance level, there is sufficient evidence to support the claim that .these sample pulse rates come from a population with a mean less than 72 beats per minute.

Use Z-Test
load data in L1

Determine whether the hypothesis test involves a sampling distribution of means that is a normal distribution, Student t distribution, or neither.

Claim: μ = 119.
Sample data: n = 11, x = 110, s = 15.2.
The sample data appear to come from a normally distributed population with unknown μ and σ.

A) Normal
B) Student t
C) Neither

B) Student t

*unknown σ

Assume that a simple random sample has been selected from a normally distributed population. Find the test statistic, P-value, critical value(s), and state the final conclusion.

Test the claim that for the population of female college students, the mean weight is given by μ = 132 lb. Sample data are summarized as n = 20, x = 137 lb, and s = 14.2 lb. Use a significance level of α = 0.1.

α = 0.1
Test statistic: t = 1.57
P-value: p = 0.1318
Critical values: t = ±1.729
Because the test statistic, t < 1.729, we fail to reject the null hypothesis.
There is not sufficient evidence to warrant rejection of the claim that μ = 132 lb.

Assume that a simple random sample has been selected from a normally distributed population and test the given claim.
Identify the null and alternative hypotheses, test statistic, critical value(s) or P-value (or range of P-values) as appropriate, and state the final conclusion that addresses the original claim.

A large software company gives job applicants a test of programming ability and the mean for that test has been 160 in the past. Twenty-five job applicants are randomly selected from one large university and they produce a mean score and standard deviation of 183 and 12, respectively. Use a 0.05 level of significance to test the claim that this sample comes from a population with a mean score greater than 160. Use the P-value method of testing hypotheses.

H0: μ = 160
H1: μ > 160
Test statistic: t = 9.583
P-value < 0.005
Reject H0
There is sufficient evidence to support the claim that the mean is greater than 160.

Assume that a simple random sample has been selected from a normally distributed population and test the given claim.
Identify the null and alternative hypotheses, test statistic, critical value(s) or P-value (or range of P-values) as appropriate, and state the final conclusion that addresses the original claim.

A cereal company claims that the mean weight of the cereal in its packets is 14 oz. The weights (in ounces) of the cereal in a random sample of 8 of its cereal packets are listed below.

14.6 13.8 14.1 13.7 14.0 14.4 13.6 14.2

Test the claim at the 0.01 significance level.

H0: μ = 14 oz.
H1: μ  14 oz.
Test statistic: t = 0.408
Critical values: t = ±3.499.
Fail to reject H0.
There is not sufficient evidence to warrant rejection of the claim that the mean weight is 14 ounces.

Assume that you plan to use a significance level of α = 0.05 to test the claim that p1 = p2, Use the given sample sizes and numbers of successes to find the pooled estimate p.

n1 = 100 n2 = 100
x1 = 33 x2 = 36

A) 0.241
B) 0.380
C) 0.310
D) 0.345

D) 0.345

2-PropZTest

A researcher finds that of 1000 people who said that they attend a religious service at least once a week, 31 stopped to help a person with car trouble. Of 1200 people interviewed who had not attended a religious service at least once a month, 22 stopped to help a person with car trouble. At the 0.05 significance level, test the claim that the two proportions are equal.

H0: p1 = p2
H1: p1  p2
Test statistic: z = 1.93
Critical values: z = ±1.96
Fail to reject the null hypothesis.
There is not sufficient evidence to warrant rejection of the claim that the two proportions are equal.

2-PropZTest
two-tailed test Critical values = (.05 significance level)/2 before InvNorm

Seven of 8500 people vaccinated against a certain disease later developed the disease. 18 of 10,000 people vaccinated with a placebo later developed the disease. Test the claim that the vaccine is effective in lowering the incidence of the disease. Use a significance level of 0.02.

H0: p1 = p2
H1: p1 < p2
Test statistic: z = -1.80.
Critical value: z = -2.05.
Fail to reject the null hypothesis.
There is not sufficient evidence to support the claim that the vaccine is effective in lowering the incidence of the disease.

2-PropZTest
Critical values = InvNorm

In a random sample of 300 women, 50% favored stricter gun control legislation. In a random sample of 200 men, 28% favored stricter gun control legislation. Construct a 98% confidence interval for the difference between the population proportions p1 - p2.

A) 0.109 < p1 - p2 < 0.331
B) 0.136 < p1 - p2 < 0.304
C) 0.120 < p1 - p2 < 0.320
D) 0.132 < p1 - p2 < 0.308

C) 0.120 < p1 - p2 < 0.320

2-PropZInt

Test the indicated claim about the means of two populations. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Use the traditional method or P-value method as indicated.

A researcher wishes to determine whether people with high blood pressure can reduce their blood pressure, measured in mm Hg, by following a particular diet. Use a significance level of 0.01 to test the claim that the treatment group is from a population with a smaller mean than the control group. Use the traditional method of hypothesis testing.

H0: μ1 = μ2.
H1: μ1 < μ2.
Test statistic: t = -8.426
Critical value: -2.364
Reject the null hypothesis.
There is sufficient evidence to support the claim that the treatment group is from a population with a smaller mean than the control group.

2-SampTTest
Critical value = InvNorm

Test the indicated claim about the means of two populations. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Use the traditional method or P-value method as indicated.

A researcher wishes to determine whether the blood pressure of vegetarians is, on average, lower than the blood pressure of nonvegetarians. Independent simple random samples of 85 vegetarians and 75 nonvegetarians yielded the following sample statistics for systolic blood pressure

Use a significance level of 0.01 to test the claim that the mean systolic blood pressure of vegetarians is lower than the mean systolic blood pressure of nonvegetarians. Use the P-value method of hypothesis testing.

H0: μ1 = μ2
H1: μ1 < μ2
Test statistic: t = -2.365
0.005 < P-value < 0.01??????
Reject H0
At the 1% significance level, there is sufficient evidence to support the claim that the mean systolic blood pressure of vegetarians is lower than the mean systolic blood pressure of nonvegetarians.

2-SampTTest
Critical value = InvNorm

Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal.

Independent samples from two different populations yield the following data. x1 = 260, x2 = 314, s1 = 75, s2 = 33. The sample size is 399 for both samples. Find the 85% confidence interval for μ1 - μ2.

A) -55 < μ1 - μ2 < -53
B) -60 < μ1 - μ2 < -48
C) -70 < μ1 - μ2 < -38
D) -62 < μ1 - μ2 < -46

D) -62 < μ1 - μ2 < -46

I keep getting B) when using 2-SampTint
B) -60 < μ1 - μ2 < -48

If anyone knows what I am doing wrong please comment below

A researcher was interested in comparing the amount of time spent watching television by women and by men. Independent simple random samples of 14 women and 17 men were selected, and each person was asked how many hours he or she had watched television during the previous week. The summary statistics are as follows.

The following 99% confidence interval was obtained for μ1 - μ2, the difference between the mean amount of time spent watching television for women and the mean amount of time spent watching television for men: -7.33 hrs < μ1 - μ2 < 2.53 hrs.

What does the confidence interval suggest about the population means?

A) The confidence interval includes only negative values which suggests that the mean amount of time spent watching television for women is smaller than the mean amount of time spent watching television for men.

B) The confidence interval limits include 0 which suggests that the two population means are unlikely to be equal. There appears to be a significant difference between the mean amount of time spent watching television for women and the mean amount of time spent watching television for men.

C) The confidence interval includes only positive values which suggests that the mean amount of time spent watching television for women is larger than the mean amount of time spent watching television for men.

D) The confidence interval limits include 0 which suggests that the two population means might be equal. There does not appear to be a significant difference between the mean amount of time spent watching television for women and the mean amount of time spent watching television for men.

D) The confidence interval limits include 0 which suggests that the two population means might be equal. There does not appear to be a significant difference between the mean amount of time spent watching television for women and the mean amount of time spent watching television for men.

Perform the indicated hypothesis test. Assume that the two samples are independent simple random samples selected from normally distributed populations. Also assume that the population standard deviations are equal (σ1 = σ2), so that the standard error of the difference between means is obtained by pooling the sample variances .

A researcher was interested in comparing the amount of time spent watching television by women and by men. Independent simple random samples of 14 women and 17 men were selected, and each person was asked how many hours he or she had watched television during the previous week. The summary statistics are as follows.

Use a 0.05 significance level to test the claim that the mean amount of time spent watching television by women is smaller than the mean amount of time spent watching television by men. Use the traditional method of hypothesis testing.

H0: μ1 = μ2
H1: μ1 < μ2
Test statistic: t = -3.451
Critical value: t = -1.699
Reject H0.
At the 5% significance level, there is sufficient evidence to support the claim that the mean amount of time spent watching television by women is smaller than the mean amount of time spent watching television by men.

2-SampTTest

Perform the indicated hypothesis test. Assume that the two samples are independent simple random samples selected from normally distributed populations. Also assume that the population standard deviations are equal (σ1 = σ2), so that the standard error of the difference between means is obtained by pooling the sample variances .

A researcher was interested in comparing the GPAs of students at two different colleges. Independent simple random samples of 8 students from college A and 13 students from college B yielded the following GPAs.

Use a 0.10 significance level to test the claim that the mean GPA of students at college A is equal to the mean GPA of students at college B. Use the traditional method of hypothesis testing.
(Note: x1 = 3.1125, x2 = 3.4385, s1 = 0.4357, s2 = 0.5485.)

H0: μ1 = μ2
H1: μ1  μ2
Test statistic: t = -1.423
Critical values: t = ±1.729
Do not reject H0.
At the 10% significance level, there is not sufficient evidence to warrant rejection of the claim that the mean GPA of students at college A is equal to the mean GPA of students at college B.

2-SampTTest

Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Also assume that the population standard deviations are equal (σ1 = σ2), so that the standard error of the difference between means is obtained by pooling the sample variances .

A researcher was interested in comparing the amount of time spent watching television by women and by men. Independent simple random samples of 14 women and 17 men were selected and each person was asked how many hours he or she had watched television during the previous week. The summary statistics are as follows.

Construct a 95% confidence interval for μ1 - μ2, the difference between the mean amount of time spent watching television for women and the mean amount of time spent watching television for men.

A) -7.30 hrs < μ1 - μ2 < -0.50 hrs
B) -6.62 hrs < μ1 - μ2 < -1.18 hrs
C) -7.18 hrs < μ1 - μ2 < -0.62 hrs
D) -7.45 hrs < μ1 - μ2 < -0.35 hrs

C) -7.18 hrs < μ1 - μ2 < -0.62 hrs

Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Also assume that the population standard deviations are equal (σ1 = σ2), so that the standard error of the difference between means is obtained by pooling the sample variances .

A paint manufacturer wanted to compare the drying times of two different types of paint. Independent simple random samples of 11 cans of type A and 9 cans of type B were selected and applied to similar surfaces. The drying times, in hours, were recorded. The summary statistics are as follows.

Construct a 99% confidence interval for μ1 - μ2, the difference between the mean drying time for paint type A and the mean drying time for paint type B.

A) -1.85 hrs < μ1 - μ2 < 8.25 hrs
B) -0.67 hrs < μ1 - μ2 < 7.07 hrs
C) 0.17 hrs < μ1 - μ2 < 6.23 hours
D) -1.16 hrs < μ1 - μ2 < 7.56 hrs

D) -1.16 hrs < μ1 - μ2 < 7.56 hrs

The two data sets are dependent. Find d to the nearest tenth.

A) 43.0
B) 20.6
C) 34.4
D) 44.7

C) 34.4

Find sd.
The differences between two sets of dependent data are 15, 27, 3, 3, 12. Round to the nearest tenth.

A) 12.9
B) 19.8
C) 7.9
D) 9.9

D) 9.9

A coach uses a new technique in training middle distance runners. The times for 9 different athletes to run 800 meters before and after this training are shown below.

Construct a 99% confidence interval for the mean difference of the "before" minus "after" times.

A) -0.85 < μd < 3.29
B) -0.76 < μd < 3.20
C) -0.82 < μd < 3.26
D) -0.54 < μd < 2.98

C) -0.82 < μd < 3.26

A test of abstract reasoning is given to a random sample of students before and after they completed a formal logic course. The results are given below. Construct a 95% confidence interval for the mean difference between the before and after scores.

A) 0.2 < μd < 7.2
B) 1.2 < μd < 5.7
C) 0.8 < μd < 6.6
D) 1.0 < μd < 6.4

A) 0.2 < μd < 7.2

Use the traditional method of hypothesis testing to test the given claim about the means of two populations. Assume that two dependent samples have been randomly selected from normally distributed populations.

Five students took a math test before and after tutoring. Their scores were as follows.

Using a 0.01 level of significance, test the claim that the tutoring has an effect on the math scores.

H0: μd = 0.
H1: μd  0.
Test statistic: t = -2.134.
Critical values: t = 4.604, -4.604.
Fail to reject H0.
There is not sufficient evidence to support the claim that the tutoring has an effect.

Use the traditional method of hypothesis testing to test the given claim about the means of two populations. Assume that two dependent samples have been randomly selected from normally distributed populations.

A coach uses a new technique to train gymnasts. 7 gymnasts were randomly selected and their competition scores were recorded before and after the training. The results are shown below.

Using a 0.01 level of significance, test the claim that the training technique is effective in raising the gymnasts' scores.

H0: μd = 0.
H1: μd < 0
Test statistic t = -0.880.
Critical value: t = -3.143.
Fail to reject H0.
There is not sufficient evidence to support the claim that the technique is effective in raising the gymnasts' scores.

Given the linear correlation coefficient r and the sample size n, determine the critical values of r and use your finding to state whether or not the given r represents a significant linear correlation. Use a significance level of 0.05.

r = 0.41, n = 25

A) Critical values: r = ±0.487, no significant linear correlation
B) Critical values: r = ±0.396, significant linear correlation
C) Critical values: r = ±0.487, significant linear correlation
D) Critical values: r = ±0.396, no significant linear correlation

B) Critical values: r = ±0.396, significant linear correlation

Which shows the strongest linear correlation?

Find the value of the linear correlation coefficient r.

A) -0.209
B) 0
C) 0.209
D) 0.186

C) 0.209

Find the value of the linear correlation coefficient r.

The paired data below consist of the test scores of 6 randomly selected students and the number of hours they studied for the test.

A) 0.224
B) 0.678
C) -0.224
D) -0.678

A) 0.224

Suppose you will perform a test to determine whether there is sufficient evidence to support a claim of a linear correlation between two variables. Find the critical values of r given the number of pairs of data n and the significance level α.

n = 14, α = 0.01

A) r = ±0.661
B) r = ±0.532
C) r = 0.661
D) r = 0.684

A) r = ±0.661

Describe the error in the stated conclusion.

Given: There is a significant linear correlation between the number of homicides in a town and the number of movie theaters in a town.

Conclusion: Building more movie theaters will cause the homicide rate to rise.

Significant correlation does not imply causality. Both variables are affected by a third variable (a lurking variable), namely the population of the town.

Describe the error in the stated conclusion.

Given: Each school in a state reports the average SAT score of its students. There is a significant linear correlation between the average SAT score of a school and the average annual income in the district in which the school is located.

Conclusion: There is a significant linear correlation between individual SAT scores and family income.

Averages suppress individual variation and tend to inflate the correlation coefficient. The fact that there is significant linear correlation between average SAT scores and average incomes in the district does not necessarily imply that there is significant linear correlation between individual SAT scores and family incomes.

Describe the error in the stated conclusion.

Given: The linear correlation coefficient between scores on a math test and scores on a test of athletic ability is negative and close to zero.

Conclusion: People who score high on the math test tend to score lower on the test of athletic ability.

Because the linear correlation coefficient is close to zero and is probably not significant, no conclusion can be reached regarding the relationship between scores on the math test and scores on the test of athletic ability.

Use the given data to find the best predicted value of the response variable.

Six pairs of data yield r = 0.444 and the regression equation yhat = 5x + 2. Also, ybar = 18.3. What is the best predicted value of y for x = 5?

A) 27
B) 93.5
C) 4.22
D) 18.3

D) 18.3

Use the given data to find the best predicted value of the response variable.

The regression equation relating attitude rating (x) and job performance rating (y) for the employees of a company is yhat = 11.7 + 1.02x. Ten pairs of data were used to obtain the equation. The same data yield r = 0.863 and ybar = 80.1. What is the best predicted job performance rating for a person whose attitude rating is 68?

A) 80.1
B) 12.6
C) 81.1
D) 79.9

C) 81.1

Use the given data to find the equation of the regression line. Round the final values to three significant digits, if necessary.

A) yhat = 0.15 + 3.0x
B) yhat = 3.0x
C) yhat = 0.15 + 2.8x
D) yhat = 2.8x

B) yhat = 3.0x

Use the given data to find the equation of the regression line. Round the final values to three significant digits, if necessary.

Ten students in a graduate program were randomly selected. Their grade point averages (GPAs) when they entered the program were between 3.5 and 4.0. The following data were obtained regarding their GPAs on entering the program versus their current GPAs.

A) yhat = 5.81 + 0.497x
B) yhat = 2.51 + 0.329x
C) yhat = 4.91 + 0.0212x
D) yhat = 3.67 + 0.0313x

D) yhat = 3.67 + 0.0313x

Use the given data to find the equation of the regression line. Round the final values to three significant digits, if necessary.

Managers rate employees according to job performance and attitude. The results for several randomly selected employees are given below.

A) yhat = 92.3 - 0.669x
B) yhat = 2.81 + 1.35x
C) yhat = -47.3 + 2.02x
D) yhat = 11.7 + 1.02x

D) yhat = 11.7 + 1.02x

Is the data point, P, an outlier, an influential point, both, or neither?

A) Both
B) Influential point
C) Neither
D) Outlier

A) Both

Is the data point, P, an outlier, an influential point, both, or neither?

A) Neither
B) Both
C) Outlier
D) Influential point

C) Outlier

Nine adults were selected at random from among those working full time in the town of Workington.
Each person was asked the number of years of college education they had completed and was also asked to rate their job satisfaction on a scale of 1 to 10. The pairs of data values area plotted in the scatterplot below.

The four points in the lower left corner correspond to employees from company A and the five points in the upper right corner correspond to employees from company B.
a. Using the pairs of values for all 9 points, find the equation of the regression line.
b. Using only the pairs of values for the four points in the lower left corner, find the equation of the regression line.
c. Using only the pairs of values for the five points in the upper right corner, find the equation of the regression line.
d. Compare the results from parts a, b, and c.

a. yhat = 0.833 + 1.25x
b. yhat = 1.5 + 0.5x
c. yhat = 10.29 - 0.643x
d. The results are very different indicating that combinations of clusters can produce results that differ dramatically from results within each cluster alone.

The following residual plot is obtained after a regression equation is determined for a set of data. Does the residual plot suggest that the regression equation is a bad model? Why or why not?

No, the residual plot does not suggest that the regression equation is a bad model. The residual plot does not have an obvious pattern that is not a straight line. This confirms that a scatterplot of the sample data is a straight line. The residual plot does not become thicker or thinner when viewed from left to right. This confirms that for different fixed values of x, the distributions of the corresponding y-values all have the same standard deviation.