Find the critical value z _{α /2} that corresponds to the given confidence level.

92%

z _{α /2} = 1.75

8% ÷ 2 = 4%

1.0000 – 0.0400 = 0.9600

Find the critical value z _{α /2} that corresponds to α = 0.04.

z _{α /2} = 2.05

0.04 ÷ 2 = 0.02

1.0000 – 0.0200 = 0.9800

Find the critical value z _{α /2} that corresponds to a 98% confidence level.

2.33

2% ÷ 2 = 1%

1.0000 – 0.0100 = 0.9900

Find the critical value z _{α /2} that corresponds to a 90% confidence level.

1.645

10% ÷ 2 = 5%

1.0000 – 0.0500 = 0.9500

Find the critical value z _{α /2} that corresponds to a 91.28% confidence level.

1.71

8.72% ÷ 2 = 4.36%

1.0000 – 0.0436 = 0.9564

Find z _{α /2} for α = 0.07

1.81

0.07 ÷ 2 = 0.035

1.0000 – 0.035 = 0.9650 ≈ 0.9649

Confidence level 95​%; n = 15​; σ is known​; population appears to be very skewed.

Do one of the​ following, as appropriate.

(a) Find the critical value z _{α /2}

(b) Find the critical value t _{α /2}

(c) State that neither the normal nor the t distribution applies.

Neither normal nor t distribution applies.

90%; n = 200​; σ = 13.0​; population appears to be skewed

Do one of the​ following, as appropriate.

(a) Find the critical value z _{α /2}

(b) Find the critical value t _{α /2}

(c) State that neither the normal nor the t distribution applies.

z_α/2 = 1.645

10% ÷ 2 = 5%

1.0000 – 0.0500 = 0.9500

Confidence level 95%; n = 19​; σ is unknown​; population appears to be normally distributed.

Do one of the​ following, as appropriate.

(a) Find the critical value z _{α /2}

(b) Find the critical value t _{α /2}

(c) State that neither the normal nor the t distribution applies.

t _α/2 = 2.101

5% ÷ 2 = 2.5%

18 Degrees of Freedom; Area in One Tail; 0.025

Confidence level 99%; n = 28​; σ = 31.6​; population appears to be normally distributed.

Do one of the​ following, as appropriate.

(a) Find the critical value z _{α /2}

(b) Find the critical value t _{α /2}

(c) State that neither the normal nor the t distribution applies.

z_α/2 = 2.575

1% ÷ 2 = 0.5%

1.0000 – 0.0050 = 0.9950

A newspaper provided a​ "snapshot" illustrating poll results from 1910 professionals who interview job applicants. The illustration showed that​ 26% of them said the biggest interview turnoff is that the applicant did not make an effort to learn about the job or the company. The margin of error was given as ±3 percentage points.

What important feature of the poll was​ omitted?

The confidence level

In this​ poll, the sample size is given as 1910​ professionals, the point estimate is given as​ 26%, and the confidence interval is given as ±3 percentage points around the point estimate.​ However, the confidence level is not provided. It is often​ 95%, but media reports often neglect to identify it.

Find the critical value z _{α /2} for α = 0.09.

z _{α /2} = 1.70

Express the confidence interval (0.043, 0.095) in the form of p̂ – E < p < p̂ + E

0.043 < p < 0.095

A research institute poll asked respondents if they acted to annoy a bad driver. In the​ poll, n = 2356, and x = 909 who said that they honked. Use a 95% confidence level.

1. Find the best point estimate of the population proportion p.
2. Identify the value of the margin of error E.
3. Construct the confidence interval.
4. Write a statement that correctly interprets the confidence interval.

a. 0.386

p̂ = x/n

= 909/2356

= 0.3858234295

b. E = 0.0197

z_α/2 = z_0.05/2 = z_0.025 = 1.96

E = z_α/2 × √(p̂×q^ ÷ n)

= 1.96 × √[0.386(1 – 0.386) ÷ 2356]

= 0.0196583139

c. 0.366 < p < 0.406

First check that the requirements to construct a confidence interval used to estimate a population proportion are met.

1. The sample is a simple random sample.

2. The conditions for the binomial distribution are satisfied.

3. There are at least 5 successes and at least 5 failures.

p̂ – E = 0.386 – 0.0197 = 0.3663

p̂ + E = 0.386 + 0.0197 = 0.4057

d. One has 95​% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.

In the week before and the week after a​ holiday, there were 10,000 total​ deaths, and 4939 of them occurred in the week before the holiday.

1. Construct a 95​% confidence interval estimate of the proportion of deaths in the week before the holiday to the total deaths in the week before and the week after the holiday.
2. Based on the​ result, does there appear to be any indication that people can temporarily postpone their death to survive the​ holiday?

a. 0.484 < p < 0.504

b. No, because the proportion could easily equal 0.5. The interval is not less than 0.5 the week before the holiday.

An online site presented this​ question, 'Would the recent norovirus outbreak deter you from taking a​ cruise?' Among the 34,742 people who​ responded, 61​% answered​ 'yes'.

Use the sample data to construct a 90​% confidence interval estimate for the proportion of the population of all people who would respond​ 'yes' to that question.

Does the confidence interval provide a good estimate of the population​ proportion?

0.606 < p < 0.614

p̂ = 0.61

z _α/2 = z_0.05 = 1.645

E = z_α/2 × √(p̂×q^ ÷ n)

= 1.645 × √[0.61(0.39) ÷ 34,742]

= 0.0043

p = 0.61 ± 0.0043

No, the sample is a voluntary sample and might not be representative of the population.

Use the data in the table above to answer the following questions.

Find the sample proportion of candy that are red.

Use that result to construct a 90% confidence interval estimate of the population percentage of candy that are red.

Is the result consistent with the 30 % rate that is reported by the candy​ maker?

The proportion of red candy = 0.25

Number of red candy = 9

Pieces of candy in sample bag = 36

The proportion of red candy = 9/36 = 0.25

13.1% < p < 36.9 %

z _α/2 = z_0.05 = 1.645

E = z_α/2 × √(p̂×q^ ÷ n)

= 1.645 × √[0.25(0.75) ÷ 36]

= 0.1187

p = 0.25 ± 0.1187 = 0.131, 0.369

Yes​, because the confidence interval includes 30​%.

A study of 420,052 cell phone users found that 139 of them developed cancer of the brain or nervous system. Prior to this study of cell phone​ use, the rate of such cancer was found to be 0.0365​% for those not using cell phones.

1. Use the sample data to construct a 95​% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system.
2. Do cell phone users appear to have a rate of cancer of the brain or nervous system that is different from the rate of such cancer among those not using cell​ phones? Why or why​ not?

a. 0.028% < p < 0.039 %

p̂ = 139 ÷ 420,052 = 0.00033091

z_0.025 = 1.96

E = 1.96 × √[0.00033091(0.99966909) ÷ 420,052] = 0.000055003

0.00033091 ± 0.000055 = 0.00028, 0.00039

b. No, because 0.0365​% is included in the confidence interval.

Many states are carefully considering steps that would help them collect sales taxes on items purchased through the Internet.

How many randomly selected sales transactions must be surveyed to determine the percentage that transpired over the​ Internet? Assume that we want to be 95​% confident that the sample percentage is within eight percentage points of the true population percentage for all sales transactions.

n = 151

z _α/2 = z_0.025 = 1.96

n = ((z _α/2 )² × 0.25) ÷ E²

= (1.96² × 0.25) ÷ 0.08²

= 150.0625

Find the sample​ size, n, needed to estimate the percentage of adults who have consulted fortune tellers. Use a 0.05 margin of​ error, use a confidence level of 98​%, and use results from a prior poll suggesting that 13​% of adults have consulted fortune tellers.

n = 245

z _α/2 = z_0.01 = 2.326

n = ((z _α/2 )² × p̂ × q^) ÷ E²

= (2.326² × 0.13 × 0.87) ÷ 0.05²

= 244.7608862

A programmer plans to develop a new software system. In planning for the operating system that he will​ use, he needs to estimate the percentage of computers that use a new operating system. How many computers must be surveyed in order to be 90​% confident that his estimate is in error by no more than four percentage points?

1. Assume that nothing is known about the percentage of computers with new operating systems.
2. Assume that a recent survey suggests that about 97​% of computers use a new operating system.
3. Does the additional survey information from part​ (b) have much of an effect on the sample size that is​ required?

a. n = 423

b. n = 50

c. Yes, using the additional survey information from part​ (b) dramatically reduces the sample size.

The required sample size decreases dramatically from 423 to 50​, with the addition of the survey information.

Which of the following groups has terms that can be used interchangeably with the​ others?

Percentage, Probability, and Proportion

​Percentage, probability, and proportion can be used interchangeably with each other. The critical value cannot be used interchangeably with these terms because the critical value is a number separating sample statistics that are likely to occur from those that are unlikely to occur.

Which of the following is NOT true of the confidence level of a confidence​ interval?

1. The confidence level is often expressed as the probability or area 1 – α, where α is the complement of the confidence level.
2. The confidence level is also called the degree of confidence.
3. There is a 1 – ​α chance, where α is the complement of the confidence​ level, that the true value of p will fall in the confidence interval produced from our sample.
4. The confidence level gives us the success rate of the procedure used to construct the confidence interval.

There is a 1 – ​α chance, where α is the complement of the confidence​ level, that the true value of p will fall in the confidence interval produced from our sample.

The confidence level is the probability that the confidence interval actually does contain the true value of​ p, not the other way around. Saying that​ "there is a 1 – α ​chance, where α is the complement of the confidence​ level, that the true value of p will fall in the confidence interval produced from our​ sample" is a common misinterpretation of the confidence interval.

Which of the following is NOT a requirement for constructing a confidence interval for estimating the population​ proportion?

1. There are a fixed number of trials.
2. There are at least 5 successes and 5 failures.
3. The trials are done without replacement.
4. The sample is a simple random sample.

The trials are done without replacement.

The trials are done without replacement is not a requirement because The​ 5% Guideline for Cumbersome Calculations states that if calculations are cumbersome and if a sample size is no more than​ 5% of the size of the​ population, treat the selections as being independent​ (even if the selections are made without​ replacement, so that they are technically​ dependent).

1. What does it mean to say that the confidence interval methods of this section are robust against departures from​ normality?
2. Does the given dotplot appear to satisfy these​ requirements?

a. The confidence interval methods of this section are robust against departures from​ normality, meaning they work well with distributions that​ aren't normal, provided that departures from normality are not too extreme.

Confidence interval methods are robust against departures from normality if either the sample size is greater than​ 30, the population is normally​ distributed, or the departure from normality is not too​ extreme, which can be checked by using a histogram or a dotplot.

b. Yes, because the dotplot resembles a normal distribution and the sample size is greater than 30.

Examine the dotplot and determine if it meets the conditions for a confidence interval robust against departures from normality. Confidence interval methods are robust against departures from normality if either the sample size is greater than​ 30, the population is normally​ distributed, or the departure from normality is not too​ extreme, which can be checked by using a histogram or a dotplot.

Using the simple random sample of weights of women from a data​ set, we obtain these sample​ statistics: n = 40 and x̄ = 147.72 lb. Research from other sources suggests that the population of weights of women has a standard deviation given by σ = 32.99 lb.

1. Find the best point estimate of the mean weight of all women.
2. Find a 99​% confidence interval estimate of the mean weight of all women.

a. The best point estimate is 147.72 lb.

b. The 99% confidence interval estimate is 134.29 lb < µ < 161.15 lb.

z_0.005 = 2.575

E = z _α/2 × σ/√(n)

= 2.575 × 32.99/√(40)

= 13.43165578

x̄ – E < µ < x̄ + E

147.72 – 13.432 < µ < 147.72 + 13.432

134.288 < µ < 161.152

Randomly selected students participated in an experiment to test their ability to determine when one minute​ (or sixty​ seconds) has passed. Forty students yielded a sample mean of 59.3 seconds.

Assuming that σ = 10.5 ​seconds, construct a 99​% confidence interval estimate of the population mean of all students.

Based on the​ result, is it likely that the​ students' estimates have a mean that is reasonably close to sixty​ seconds?

The 99% confidence interval for the population mean is 55.0 < µ < 63.6.

z_0.005 = 2.575

E = z _α/2 × σ/√(n)

= 2.575 × 10.5/√(40)

= 4.275004112

x̄ – E < µ < x̄ + E

59.3 – 4.275 < µ < 59.3 + 4.275

55.025 < µ < 63.575

Yes, because the confidence interval includes sixty seconds.

A study of the ages of motorcyclists killed in crashes involves the random selection of 145 drivers with a mean of 31.66 years.

Assuming that σ = 10.7 ​years, construct a 95​% confidence interval estimate of the mean age of all motorcyclists killed in crashes.

Notice that the confidence interval limits do not include ages below 20 years. What does this​ mean?

The 95% confidence interval for the population mean is 29.92 < µ < 33.40.

z_0.025 = 1.96

E = z _α/2 × σ/√(n)

= 1.96 × 10.7/√(145)

= 1.741629803

x̄ – E < µ < x̄ + E

31.66 – 1.7416 < µ < 31.66 + 1.7416

29.9184 < µ < 33.4016

The mean age of the population will most likely not be less than 20 years old.

Salaries of 38 college graduates who took a statistics course in college have a​ mean, x̄, of $69,000.

Assuming a standard​ deviation, σ​, of ​$15,315​, construct a 99​% confidence interval for estimating the population mean mu.

$ 62,603 < µ < $ 75,397

Confidence level 99​%; n = 24​; σ is known​; population appears to be very skewed.

Do one of the​ following, as appropriate.

(a) Find the critical value z _{α /2}

(b) Find the critical value t _{α /2}

(c) State that neither the normal nor the t distribution applies.

Neither normal nor t distribution applies.

Confidence level 98​%; n = 27​; σ is unknown​; population appears to be normally distributed.

Do one of the​ following, as appropriate.

(a) Find the critical value z _{α /2}

(b) Find the critical value t _{α /2}

(c) State that neither the normal nor the t distribution applies.

t _α/2 = 2.479

Confidence level 99​%; n = 28​; σ is known​; population appears to be very skewed.

Do one of the​ following, as appropriate.

(a) Find the critical value z _{α /2}

(b) Find the critical value t _{α /2}

(c) State that neither the normal nor the t distribution applies.

Neither normal nor t distribution applies.

Confidence level 99%; n = 20​; σ is unknown​; population appears to be normally distributed.

Do one of the​ following, as appropriate.

(a) Find the critical value z _{α /2}

(b) Find the critical value t _{α /2}

(c) State that neither the normal nor the t distribution applies.

t _α/2 = 2.861

Find the critical value z _α/2 that corresponds to a​ 98% confidence level.

2.33

Find the critical value –z _α/2 that corresponds to a​ confidence level of 97.8%.

–2.29

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

Do one of the​ following, as appropriate.

(a) Find the critical value z _{α /2}

(b) Find the critical value t _{α /2}

(c) State that neither the normal nor the t distribution applies.

z _α/2 = 2.33

​90%; n = ​10; σ is unknown; population appears to be normally distributed.

Do one of the​ following, as appropriate.

(a) Find the critical value z _{α /2}

(b) Find the critical value t _{α /2}

(c) State that neither the normal nor the t distribution applies.

t _α/2 = 1.83

​90%; n = ​9; σ = 4.2; population appears to be very skewed.

Do one of the​ following, as appropriate.

(a) Find the critical value z _{α /2}

(b) Find the critical value t _{α /2}

(c) State that neither the normal nor the t distribution applies.

Neither the normal nor the t distribution applies.

Weight lost on a diet: 90% confidence n = 51 x̄ = 3.0 kg s = 5.1 kg

Use technology and the given confidence level and sample data to find the confidence interval for the population mean µ. Assume that the population does not exhibit a normal distribution.

Is the confidence interval affected by the fact that the data appear to be from a population that is not normally​ distributed?

1.8 kg < µ < 4.2 kg

t _α/2 = t_0.05 = 1.676

E = t _α/2 × s÷√(n)

= 1.676 × 5.1÷√(51)

= 1.20

x̄ – E < µ < x̄ + E

3.0 – 1.2 < µ < 3.0 + 1.2

1.8 < µ < 4.2

No, because the sample size is large enough.

Listed below are measured amounts of lead​ (in micrograms per cubic​ meter, or µg/m³​) in the air. The EPA has established an air quality standard for lead of 1.5 µg/m³. The measurements shown below were recorded at a building on different days.

5.40 1.10 0.47 0.75 0.71 1.30

Use the given values to construct a 95​% confidence interval estimate of the mean amount of lead in the air.

Is there anything about this data set suggesting that the confidence interval might not be very​ good?

–0.345 µg/m³ < µ < 3.589 µg/m³

(calc): s = 1.874; x̄ = 1.622

t_{df, α/2} = t_{5, 0.025} = 2.571

E = t _α/2 × s÷√(n)

= 2.571 × 1.874÷√(6)

= 1.967

x̄ – E < µ < x̄ + E

1.622 – 1.967 < µ < 1.622 + 1.967

-0.345 < µ < 3.589

Yes, the value of 5.40 appears to be an outlier.

In a test of the effectiveness of garlic for lowering​ cholesterol, 42 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes in their levels of LDL cholesterol​ (in mg/dL) have a mean of 3.4 and a standard deviation of 18.1.

1. What is the best point estimate of the population mean net change in LDL cholesterol after the garlic​ treatment?
2. Construct a 95​% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment.
3. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL​ cholesterol?

a. The best point estimate is 3.4 ​mg/dL.

x̄ = 3.4 mg/dL

b. –2.24 mg/dL < µ < 9.04 mg/dL

t_{df, α/2} = t_{41, 0.025} = 2.0195

E = t _α/2 × s÷√(n)

= 2.020 × 18.1÷√(42)

= 5.641639081

x̄ – E < µ < x̄ + E

3.4 – 5.642 < µ < 3.4 + 5.6442

-2.242 < µ < 9.042

c. The confidence interval limits contain ​0, suggesting that the garlic treatment did not affect the LDL cholesterol levels.

Twelve different video games showing substance use were observed and the duration times of game play​ (in seconds) are listed below. The design of the study justifies the assumption that the sample can be treated as a simple random sample.

4044 3877 3852 4017 4308 4803 4660 4028 5010 4817 4342 4313

Use the data to construct a 99​% confidence interval estimate of µ​, the mean duration of game play.

3983.0 < µ < 4695.5

(calc): s = 397.349; x̄ = 4339.25

t_{df, α/2} = t_{11, 0.005} = 3.106

E = t _α/2 × s÷√(n)

= 3.106 × 397.349÷√(12)

= 356.2730344

x̄ – E < µ < x̄ + E

4339.25 – 356.273 < µ < 4339.25 + 356.273

3982.977 < µ < 4695.523

A physician wants to develop criteria for determining whether a​ patient's pulse rate is​ atypical, and she wants to determine whether there are significant differences between males and females. Use the sample pulse rates above.

1. Construct a 95​% confidence interval estimate of the mean pulse rate for males.
2. Construct a 95​% confidence interval estimate of the mean pulse rate for females.
3. Compare the preceding results. Can we conclude that the population means formales and females are​ different?

a. 64.0 < µ < 82.4

(calc): s = 12.795; x̄ = 73.2

t_{df, α/2} = t_{9, 0.025} = 2.262

E = t _α/2 × s÷√(n)

= 2.262 × 12.795÷√(10)

= 9.15235571

x̄ – E < µ < x̄ + E

73.2 – 9.2 < µ < 73.2 + 9.2

64 < µ < 82.4

b. 64.7 < µ < 90.5

(calc): s = 18.007; x̄ = 77.6

t_{df, α/2} = t_{9, 0.025} = 2.262

E = t _α/2 × s÷√(n)

= 2.262 × 18.007÷√(10)

=12.88053687

x̄ – E < µ < x̄ + E

77.6 – 12.9 < µ < 77.6 + 12.9

64.7 < µ < 90.5

c. No, because the two confidence intervals​ overlap, we cannot conclude that the two population means are different.

An IQ test is designed so that the mean is 100 and the standard deviation is 17 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 99​% confidence that the sample mean is within 7 IQ points of the true mean.

Assume that σ = 17 and determine the required sample size using technology.

Then determine if this is a reasonable sample size for a real world calculation.

The required sample size is 40.

z _α/2 = z_0.005 = 2.575

n = [ (z _α/2 × σ) ÷ E ] ²

= [ (2.575 × 17) ÷ 7 ] ²

= 39.10715561

Yes. This number of IQ test scores is a fairly small number.

A student wants to estimate the mean score of all college students for a particular exam.

First use the range rule of thumb to make a rough estimate of the standard deviation of those scores. Possible scores range from 300 to 2200.

Use technology and the estimated standard deviation to determine the sample size corresponding to a 95​% confidence level and a margin of error of 100 points.

What​ isn't quite right with this​ exercise?

The range rule of thumb estimate for the standard deviation is 475.

σ ≈ range ÷ 4

≈ (2200 – 300) ÷ 4

≈ 475

A confidence level of 95​% requires a minimum sample size of 87.

z _α/2 = z_0.025 = 1.96

n = [ (z _α/2 × σ) ÷ E ] ²

= [ (1.96 × 475) ÷ 100 ] ²

= 86.6761

A margin of error of 100 points seems too high to provide a good estimate of the mean score.

Which of the following is NOT a property of the Student t​ distribution?

1. The Student t distribution has the same general symmetric bell shape as the standard normal​ distribution, but it reflects the greater variability that is expected with small samples.
2. The Student t distribution has a mean of t = 0.
3. The standard deviation of the Student t distribution is s = 1.
4. The Student t distribution is different for different sample sizes.

The standard deviation of the Student t distribution is s = 1.

The standard deviation of the Student t distribution is greater than​ 1, unlike the standard normal​ distribution, which has a standard deviation of 1.

Which of the following calculations is NOT derived from the confidence​ interval?

1. The point estimate of µ, x̄ = (upper + lower confidence limit) ÷ 2
2. Difference between the​ limits, 2E = (upper confidence limit) – (lower confidence limit)
3. The population​ mean, µ = (upper confidence limit) + (lower confidence limit)
4. The margin of​ error, E = (upper – lower confidence limit) ÷ 2

The population​ mean, µ = (upper confidence limit) + (lower confidence limit)

The population mean is not calculated from the confidence interval​ estimate; it is a fixed constant.

Which of the following is NOT a requirement for constructing a confidence interval for estimating a population mean with σ ​known?

1. The confidence level is​ 95%.
2. Either the population is normally distributed or n > ​30, or both.
3. The sample measures a quantitative value.
4. The sample is a simple random sample.

The confidence level is​ 95%.

A confidence level of​ 95% is not a requirement for constructing a confidence interval to estimate the mean with sigma known. The confidence level may vary.

Which of the following would be a correct interpretation of a​ 99% confidence interval such as 4.1 < µ < 5.6?

We are​ 99% confident that the interval from 4.1 to 5.6 actually does contain the true value of µ.

If we were to select many different samples of the same size and construct the corresponding confidence​ intervals, in the long​ run, 99% of them would actually contain the value of µ.

Which of the following is NOT an equivalent expression for the confidence interval given by 161.7 < µ < 89.5?

1. 175.6 – 13.9 < µ < 175.6 + 13.9
2. 161.7 ±27.8
3. (161.7,189.5)
4. 175.6 ±13.9

161.7 ±27.8

The interval 161.7 ±27.8 would actually produce 133.9 < µ < 189.5.

Which of the following is NOT required to determine minimum sample size to estimate a population​ mean?

1. The desired confidence level
2. The size of the​ population, N
3. The value of the population standard​ deviation, σ
4. The desired margin of error

The size of the​ population, N

The minimum sample size does not depend on the size of the population of interest.