Chapter 6 Review HOMEWORK
Use the values on the number line to find the sampling error. (Since I don't have Quizlet+, I can't insert the image of the actual infinite number line; ergo, I pasted the description.) An infinite number line, labeled from 3 to 5, has tick marks in increments of 0.2. From left to right, a point labeled *"x̄" = 3.6* is plotted at 3.6, and a point labeled *"μ" = 4.46* is plotted at 4.46. The sampling error is *___*.
Correct Answer: *-0.86* (*6.1.9*)
Find the margin of error for the given values of *c*, *σ*, and *n*. (Round answer to *three* decimal places.) *c = 0.90, σ = 2.5, n = 49* *Level of Confidence* = *z*∨*c* *90%* = *1.645* *95%* = *1.96* *99%* = *2.575* *E* = *____*
Correct Answer: *0.588* (*6.1.13*)
Find the margin of error for the given values of *c*, *σ*, and *n*. (Round answer to *three* decimal places.) *c = 0.95, σ = 3.5, n = 49* *Level of Confidence* = *z*∨*c* *90%* = *1.645* *95%* = *1.96* *99%* = *2.575* *E* = *___*
Correct Answer: *0.98* (*6.1.15*)
Use the values on the number line to find the sampling error. (Since I don't have Quizlet+, I can't insert the image of the actual infinite number line; ergo, I pasted the description.) An infinite number line, labeled from 25 to 28, has tick marks in increments of 0.5. Labeled points are plotted at *"μ" = 25.72* and *"x̄" = 26.87*. The sampling error is *___*.
Correct Answer: *1.15* (*6.1.11*)
Find the critical value *z*∨*c* necessary to form a confidence interval at the level of confidence shown below. (Round answer to *two* decimal places.) *"c" = 0.81* *z*∨*c* = *___*
Correct Answer: *1.31* (*6.1.5*)
Find the critical value *t*∨*c* for the confidence level c = 0.80 and sample size n = 18. (Round answer to the *nearest thousandth* (*three* decimal places).) *(I didn't copy the accompanying "t" - Distribution Table because it's too big.)* *t*∨*c* = *____*
Correct Answer: *1.333* (*6.2.1*)
Find the margin of error for the given values of *c*, *s*, and *n*. (Round answer to *three* decimal places.) *c = 0.98, s = 2.9, n = 18* *(I didn't copy the accompanying "t" - Distribution Table because it's too big.)* The margin of error is *____*.
Correct Answer: *1.755* (*6.2.7*)
Find the critical value *z*∨*c* necessary to form a confidence interval at the level of confidence shown below. (Round answer to *two* decimal places.) *"c" = 0.96* *z*∨*c* = *___*
Correct Answer: *2.05* (*6.1.7*)
Find the critical value *t*∨*c* for the confidence level c = 0.98 and sample size n = 18. (Round answer to the *nearest thousandth* (*three* decimal places).) *(I didn't copy the accompanying "t" - Distribution Table because it's too big.)* *t*∨*c* = *____*
Correct Answer: *2.567* (*6.2.3*)
Find the minimum size *n* needed to estimate *μ* for the given values of *c*, *σ*, and *E*. (Round answer *up* to the *nearest whole number* (*zero* decimal places).) *c = 0.95, σ = 5.7*, and *E = 2* Assume that a preliminary sample has at least 30 members. *n* = *__*
Correct Answer: *32* (*6.1.29*)
Find the margin of error for the given values of *c*, *s*, and *n*. (Round answer to *three* decimal places.) *c = 0.98, s = 5, n = 11* *(I didn't copy the accompanying "t" - Distribution Table because it's too big.)* The margin of error is *____*.
Correct Answer: *4.167* (*6.2.5*)
Find the minimum size *n* needed to estimate *μ* for the given values of *c*, *σ*, and *E*. (Round answer *up* to the *nearest whole number* (*zero* decimal places).) *c = 0.98, σ = 6.2*, and *E = 2* Assume that a preliminary sample has at least 30 members. *n* = *__*
Correct Answer: *53* (*6.1.31*)
Which statistic is the best unbiased estimator for *μ*? The best unbiased estimated for *μ* is *_.*
Correct Answer: *x̄.* (*6.1.2*)
When estimating a population mean, are you more likely to be correct when you use a point estimate or an interval estimate? Explain your reasoning. Choose the correct answer below. A.) You are more likely to be correct using an interval estimate because it is unlikely that a point estimate will exactly equal the population mean. B.) If *n* ≤ 30, an interval estimate is more accurate. If *n* > 30, a point estimate is more accurate. C.) You are more likely to be correct using a point estimate because an interval estimate is too broad and contains many possible values. D.) There is no difference between an interval estimate and a point estimate in terms of accuracy.
Correct Answer: A.) You are more likely to be correct using an interval estimate because it is unlikely that a point estimate will exactly equal the population mean. (*6.1.1*)
Determine whether the statement is true or false. *To estimate the value of p, the population proportion of successes, use the point estimate x.* A.) False, to estimate the value of p, use the point estimate *p̂* = *n* / *x*. B.) False, to estimate the value of p, use the point estimate *p̂* = *x* / *n*. C.) True, to estimate the value of p, use the point estimate *p̂* = *x*. D.) False, to estimate the value of p, use the point estimate *p̂* = *n* / *q̂*.
Correct Answer: B.) False, to estimate the value of p, use the point estimate *p̂* = *x* / *n*. (*6.3.1*)
Determine whether the statement is true or false. If it is false, rewrite it as a true statement. *The point estimate for the population proportion of failures is 1 − "p̂".* Choose the correct answer below. A.) The statement is false. The actual point estimate for the population proportion of failures is √(*p̂*). B.) The statement is false. The actual point estimate for the population proportion of failures is *p̂*. C.) The statement is true. D.) The statement is false. The actual point estimate for the population proportion of failures is 1 − √(*p̂*).
Correct Answer: C.) The statement is true. (*6.3.2*)
In a random sample of 7 cell phones, the mean full retail price was $518.70 and the standard deviation was $186.00. Further research suggests that the population mean is $429.11. Does the t-value for the original sample fall between *−t₀.₉₅* and *t₀.₉₅*? Assume that the population of full retail prices for cell phones is normally distributed. (Round all answers to *two* decimal places.) The t-value of t = *__(1)__* *__(2)__* fall between *−t₀.₉₅* and *t₀.₉₅* because *t₀.₉₅* = *__(3)__*.
Correct Answers: *(1):* *1.27* *(2):* *does* *(3):* *2.45* (*6.2.23-T*)
When a population is finite, the formula that determines the standard error of the mean *σ*∨*x̄* needs to be adjusted. If *N* is the size of the population and *n* is the size of the sample (where *n* ≥ 0.05*N*), then the standard error of the mean is *σ*∨*x̄* = ((*σ*)/√(*n*)) × √((*N* − *n*)/(*N* − 1)). The expression √((*N* − *n*)/(*N* − 1)) is called the finite population correction factor. Use the finite population correction factor to construct the confidence interval for the population mean described below. (Round answer(s) to *two* decimal places.) *c = 0.99* *x̄ = 59.5* *σ = 0.3* *N = 200* *n = 25* The 99% confidence interval for the population mean is (*__(1)__*, *__(2)__*).
Correct Answers: *(1):* *59.36* *(2):* *59.64* (*6.1.58-T*)
A company manufactures light bulbs. The company wants the bulbs to have a mean life span of 994 hours. This average is maintained by periodically testing random samples of 25 light bulbs. If the t-value falls between *−t₀.₉₅* and *t₀.₉₅*, then the company will be satisfied that it is manufacturing acceptable light bulbs. For a random sample, the mean life span of the sample is 999 hours and the standard deviation is 23 hours. Assume that life spans are approximately normally distributed. Is the company making acceptable light bulbs? Explain. (Round all answers to *two* decimal places.) The company *_(1)_* making acceptable light bulbs because the t-value for the sample is t = *__(2)__* and *t₀.₉₅* = *__(3)__*.
Correct Answers: *(1):* *is* *(2):* *1.09* *(3):* *2.06* (*6.2.42-T*)
When a population is finite, the formula that determines the standard error of the mean needs to be adjusted. If *N* is the size of the population and *n* is the size of the sample (where *n* ≥ 0.05*N*), the standard error of the mean is shown below. The finite population correction factor is given by the expression √((*N* − *n*)/(*N* − 1)). Determine the finite population correction factor for each of the following (round all answers to *three* decimal places). *Part 1 (a):* *N = 1,900* and *n = 1,425* *Part 2 (b):* *N = 1,900* and *n = 190* *Part 3 (c):* *N = 1,900* and *n = 175* *Part 4 (d):* *N = 1,900* and *n = 95* *Part 5 (e):* What happens to the finite population correction factor as the sample size *n* decreases, but the population size *N* remains the same? *Part 1 (a):* The finite population correction factor is *__*. *Part 2 (b):* The finite population correction factor is *____*. *Part 3 (c):* The finite population correction factor is *____*. *Part 4 (d):* The finite population correction factor is *____*. *Part 5 (e):* What happens to the finite population correction factor as the sample size *n* decreases, but the population size *N* remains the same? A.) The finite population correction factor does not approach any set value. B.) The finite population correction factor approaches 0. C.) The finite population correction factor approaches 1. D.) The finite population correction factor approaches √(1/2) ≈ 0.707.
Correct Answers: *Part 1 (a):* *0.5* *Part 2 (b):* *0.949* *Part 3 (c):* *0.964* *Part 4 (d):* *0.975* *Part 5 (e):* C.) The finite population correction factor approaches 1. (*6.1.57*)
A researcher wishes to estimate, with 99% confidence, the population proportion of adults who support labeling legislation for genetically modified organisms (GMOs). Her estimate must be accurate within 7% of the true proportion. *Part 1 (a):* No preliminary estimate is available. Find the minimum sample size needed. (Round answer *up* to the *nearest whole number* (*zero* decimal places).) *Part 2 (b):* Find the minimum sample size needed, using a prior study that found that 79% of the respondents said they supported labeling legislation for GMOs. (Round answer *up* to the *nearest whole number* (*zero* decimal places).) *Part 3 (c):* Compare the results from parts 1 (a) and 2 (b). *Part 1 (a):* What is the minimum sample size needed assuming that no prior information is available? n = *___* *Part 2 (b):* What is the minimum sample size needed using a prior study that found that 79% of the respondents support labeling legislation? n = *___* *Part 3 (c):* How do the results from parts 1 (a) and 2 (b) compare? A.) Having an estimate of the population proportion has no effect on the minimum sample size needed. B.) Having an estimate of the population proportion raises the minimum sample size needed. C.) Having an estimate of the population proportion reduces the minimum sample size needed.
Correct Answers: *Part 1 (a):* *340* *Part 2 (b):* *226* *Part 3 (c):* C.) Having an estimate of the population proportion reduces the minimum sample size needed. (*6.3.18-T*)
A company manufactures tennis balls. When its tennis balls are dropped onto a concrete surface from a height of 100 inches, the company wants the mean height the balls bounce upward to be 54.9 inches. This average is maintained by periodically testing random samples of 25 tennis balls. If the t-value falls between *−t₀.₉₈* and *t₀.₉₈*, then the company will be satisfied that it is manufacturing acceptable tennis balls. A sample of 25 balls is randomly selected and tested. The mean bounce height of the sample is 56.3 inches and the standard deviation is 0.25 inch. Assume the bounce heights are approximately normally distributed. Is the company making acceptable tennis balls? *Part 1:* Find *−t₀.₉₉* and *t₀.₉₉*. (Round answers to *three* decimal places.) *−t₀.₉₈* = *__(1)__* *t₀.₉₈* = *__(2)__* *Part 2:* Find the t-value. t-value = *__* *Part 3:* Is the company making acceptable tennis balls? Choose the correct answer below. A.) The tennis balls are acceptable because the t-value falls between *−t₀.₉₈* and *t₀.₉₈*. B.) The tennis balls are not acceptable because the t-value falls outside *−t₀.₉₈* and *t₀.₉₈*. C.) The sample is not large enough to determine if the tennis balls are acceptable.
Correct Answers: *Part 1:* *(1):* *-2.492* *(2):* *2.492* *Part 2:* *40* *Part 3:* B.) The tennis balls are not acceptable because the t-value falls outside *−t₀.₉₈* and *t₀.₉₈*. (*6.2.41-T*)
*In a survey of "2,061" adults in a recent year, "725" made a New Year's resolution to eat healthier.* Construct 90% and 95% confidence intervals for the population proportion (round answers to *three* decimal places). Interpret the results and compare the widths of the confidence intervals. *Part 1:* The 90% confidence interval for the population proportion p is (*__(1)__*, *__(2)__*). *Part 2:* The 95% confidence interval for the population proportion p is (*__(1)__*, *__(2)__*). *Part 3:* With the given confidence, it can be said that the *________(1)________* of adults who say they have made a New Year's resolution to eat healthier is *__________(2)__________* of the given confidence interval. *Part 4:* Compare the widths of the confidence intervals. Choose the correct answer below. A.) The 90% confidence interval is wider. B.) The 95% confidence interval is wider. C.) The confidence intervals are the same width. D.) The confidence intervals cannot be compared.
Correct Answers: *Part 1:* *(1):* *0.335* *(2):* *0.369* *Part 2:* *(1):* *0.331* *(2):* *0.373* *Part 3:* *(1):* *population proportion* *(2):* *between the endpoints* *Part 4:* B.) The 95% confidence interval is wider. (*6.3.12-T*)
Translate the statement into a confidence interval (round answers to *three* decimal places). Approximate the level of confidence (round answer to *one* decimal place). In a survey of *1,100 adults* in a country, *69%* think teaching is one of the most important jobs in the country today. The survey's margin of error is *±4%*. *Part 1:* The confidence interval for the proportion is (*__(1)__*, *__(2)__*). *Part 2:* The interval is a *___%* confidence interval.
Correct Answers: *Part 1:* *(1):* *0.65* *(2):* *0.73* *Part 2:* *99.6%* (*6.3.31-T*)
*In a survey of "2,045" adults in a recent year, "1,486" say they have made a New Year's resolution.* Construct 90% and 95% confidence intervals for the population proportion (round all answers to *three* decimal places). Interpret the results and compare the widths of the confidence intervals. *Part 1:* The 90% confidence interval for the population proportion p is (*__(1)__*, *__(2)__*). *Part 2:* The 95% confidence interval for the population proportion p is (*__(1)__*, *__(2)__*). *Part 3:* With the given confidence, it can be said that the *________(1)________* of adults who say they have made a New Year's resolution is *__________(2)__________* of the given confidence interval. *Part 4:* Compare the widths of the confidence intervals. Choose the correct answer below. A.) The 90% confidence interval is wider. B.) The 95% confidence interval is wider. C.) The confidence intervals are the same width. D.) The confidence intervals cannot be compared.
Correct Answers: *Part 1:* *(1):* *0.711* *(2):* *0.743* *Part 2:* *(1):* *0.708* *(2):* *0.746* *Part 3:* *(1):* *population proportion* *(2):* *between the endpoints* *Part 4:* B.) The 95% confidence interval is wider. (*6.3.11-T*)
In a random sample of seven people, the mean driving distance to work was 20.4 miles and the standard deviation was 6.8 miles. Assume the population is normally distributed and use the *t* - distribution to find the margin of error and construct a 99% confidence interval for the population mean *μ*. (Round answers to *one* decimal place.) Interpret the results. (Type answer(s) as either *integers* or *decimals*, but *DO NOT ROUND*.) *Part 1:* Identify the margin of error. *__(1)__* *___(2)___* *Part 2:* Construct a 99% confidence interval for the population mean. (*_(1)_*, *__(2)__*) *Part 3:* Interpret the results. Select the correct choice below and fill in the answer box to complete your choice. A.) With *__%* confidence, it can be said that most driving distances to work (in miles) in the population are between the interval's endpoints. B.) With *__%* confidence, it can be said that the population mean driving distance to work (in miles) is between the interval's endpoints. C.) It can be said that *__%* of the population has a driving distance to work (in miles) that is between the interval's endpoints. D.) *__%* of all random samples of seven people from the population will have a mean driving distance to work (in miles) that is between the interval's endpoints.
Correct Answers: *Part 1:* *(1):* *11.2* *(2):* *miles* *Part 2:* *(1):* *9.2* *(2):* *31.6* *Part 3:* B.) With *99%* confidence, it can be said that the population mean driving distance to work (in miles) is between the interval's endpoints. (*6.2.18-T*)
You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean (round answers to *two* decimal places). Interpret the results and compare the widths of the confidence intervals. *From a random sample of 39 business days, the mean closing price of a certain stock was $124.80. Assume the population standard deviation is $11.14.* *Part 1:* The 90% confidence interval is (*___(1)___*, *___(2)___*). *Part 2:* The 95% confidence interval is (*__(1)__*, *__(2)__*). *Part 3:* Which interval is wider? Choose the correct answer below. A.) The 90% confidence interval B.) The 95% confidence interval *Part 4:* Interpret the results. A.) You can be certain that the population mean price of the stock is either between the lower bounds of the 90% and 95% confidence intervals or the upper bounds of the 90% and 95% confidence intervals. B.) You can be 90% confident that the population mean price of the stock is between the bounds of the 90% confidence interval, and 95% confident for the 95% interval. C.) You can be certain that the closing price of the stock was within the 90% confidence interval for approximately 35 of the 39 days, and was within the 95% confidence interval for approximately 37 of the 39 days. D.) You can be 90% confident that the population mean price of the stock is outside the bounds of the 90% confidence interval, and 95% confident for the 95% interval.
Correct Answers: *Part 1:* *(1):* *121.87* *(2):* *127.73* *Part 2:* *(1):* *121.3* *(2):* *128.3* *Part 3:* B.) The 95% confidence interval *Part 4:* B.) You can be 90% confident that the population mean price of the stock is between the bounds of the 90% confidence interval, and 95% confident for the 95% interval. (*6.1.36-T*)
Use the given confidence interval to find the margin of error and the sample proportion. (Type all answers as either *integers* or *decimals*.) *(0.729, 0.755)* *Part 1:* *Ε* = *____* *Part 2:* *p̂* = *____*
Correct Answers: *Part 1:* *0.013* *Part 2:* *0.742* (*6.3.9*)
Use the given confidence interval to find the margin of error and the sample proportion. (Type all answers as either *integers* or *decimals*.) *(0.636, 0.666)* *Part 1:* *Ε* = *____* *Part 2:* *p̂* = *____*
Correct Answers: *Part 1:* *0.015* *Part 2:* *0.651* (*6.3.7*)
Use the confidence interval to find the margin of error (round answer to *two* decimal places), and the sample mean (type answer as either an *integer* or a *decimal*). *(1.63, 2.05)* *Part 1:* The margin of error is *___*. *Part 2:* The sample mean is *___*.
Correct Answers: *Part 1:* *0.21* *Part 2:* *1.84* (*6.1.27*)
Let p be the population proportion for the following condition. Find the point estimates for p and q. (Round answers to *three* decimal places.) *In a survey of "1,487" adults from "country A", "326" said that they were not confident that the food they eat in "country A" is safe.* *Part 1:* The point estimate for p, *p̂*, is *____*. *Part 2:* The point estimate for q, *q̂*, is *____*.
Correct Answers: *Part 1:* *0.219* *Part 2:* *0.781* (*6.3.3*)
Let p be the population proportion for the following condition. Find the point estimates for p and q. (Round answers to *three* decimal places.) *[In a] study of "4,603" adults from "country A", "2,856" think mainstream media is more interested in making money than in telling the truth.* *Part 1:* The point estimate for p, *p̂*, is *___*. *Part 2:* The point estimate for q, *q̂*, is *___*.
Correct Answers: *Part 1:* *0.62* *Part 2:* *0.38* (*6.3.5*)
Use the given confidence interval to find the margin of error and the sample mean. (Type answers as either *integers* or *decimals*.) *(13.5, 20.9)* *Part 1:* The sample mean is *___*. *Part 2:* The margin of error is *__*.
Correct Answers: *Part 1:* *17.2* *Part 2:* *3.7* (*6.2.13*)
Use the given confidence interval to find the margin of error and the sample mean. (Type answers as either *integers* or *decimals*.) *(3.57, 8.55)* *Part 1:* The sample mean is *___*. *Part 2:* The margin of error is *___*.
Correct Answers: *Part 1:* *6.06* *Part 2:* *2.49* (*6.2.14*)
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In a sample of 14 randomly selected high school seniors, the mean score on a standardized test was 1,194 and the standard deviation was 168.0. Further research suggests that the population mean score on this test for high school seniors is 1,010. Does the t-value for the original sample fall between *−t₀.₉₉* and *t₀.₉₉*? Assume that the population of full retail prices for cell phones is normally distributed. (Round all answers to *two* decimal places.) The t-value of t = *_(1)_* *____(2)____* fall between *−t₀.₉₉* and *t₀.₉₉* because *t₀.₉₉* = *__(3)__*.
Correct Answers: *(1):* *4.1* *(2):* *does not* *(3):* *3.01* (*6.2.29-T*)
For each level of confidence c below, determine the corresponding normal confidence interval. Assume each confidence interval is constructed for the same sample statistics. *17.)* c = 0.88 *18.)* c = 0.90 *19.)* c = 0.95 *20.)* c = 0.98 Drag (match) each normal confidence interval given above to (with) the level of confidence *c.* (Since I don't have Quizlet+, I can't insert the images of the normal confidence intervals; ergo, I pasted each of their sample means (*x̄*), and their *lower* & *upper* limits. (Also, MyMathLab doesn't show the confidence intervals *BEFORE* I completed the problem.)) 17.) 18.) 19.) 20.)
Correct Answers: *17.)* *Lower:* *56.1* *x̄* = *57.8* *Upper:* *59.5* *18.)* *Lower:* *56* *x̄* = *57.8* *Upper:* *59.6* *19.)* *Lower:* *55.6* *x̄* = *57.8* *Upper:* *60* *20.)* *Lower:* *55.2* *x̄* = *57.8* *Upper:* *60.4* (*6.1.17-20*)
An admissions director wants to estimate the mean age of all students enrolled at a college. The estimate must be within 1.3 years of the population mean. Assume the population of ages is normally distributed. *Part 1 (a):* Determine the minimum sample size required to construct a 90% confidence interval for the population mean. Assume the population standard deviation is 1.5 years. (Round answer *up* to the *nearest whole number* (*zero* decimal places).) *Part 2 (b):* The sample mean is 20 years of age. Using the minimum sample size with a 90% level of confidence, does it seem likely that the population mean could be within 9% of the sample mean? Within 10% of the sample mean? Explain. (Round all answers to *two* decimal places.) *(I didn't copy the accompanying Standard Normal Tables because they're too big.)* *Part 1 (a):* The minimum sample size required to construct a 90% confidence interval is *_* students. *Part 2 (b):* The 90% confidence interval is (*__(1)__*, *__(2)__*). It *__(3)__* likely that the population mean could be within 9% of the sample mean because the interval formed by the values 9% away from the sample mean *______(4)______* the confidence interval. It *__(5)__* likely that the population mean could be within 10% of the sample mean because the interval formed by the values 10% away from the sample mean *______(6)______* the confidence interval.
Correct Answers: *Part 1 (a):* *4* *Part 2 (b):* *(1):* *18.77* *(2):* *21.23* *(3):* *seems* *(4):* *entirely contains* *(5):* *seems* *(6):* *entirely contains* (*6.1.50*)
A researcher wishes to estimate, with 99% confidence, the population proportion of adults who think Congress is doing a good or excellent job. Her estimate must be accurate within 2% of the true proportion. *Part 1 (a):* No preliminary estimate is available. Find the minimum sample size needed. (Round answer *up* to the *nearest whole number* (*zero* decimal places).) *Part 2 (b):* Find the minimum sample size needed, using a prior study that found that 22% of the respondents said they think Congress is doing a good or excellent job. (Round answer *up* to the *nearest whole number* (*zero* decimal places).) *Part 3 (c):* Compare the results from parts 1 (a) and 2 (b). *Part 1 (a):* What is the minimum sample size needed assuming that no prior information is available? n = *____* *Part 2 (b):* What is the minimum sample size needed using a prior study that found that 22% of the respondents said they think Congress is doing a good or excellent job? n = *____* *Part 3 (c):* How do the results from parts 1 (a) and 2 (b) compare? A.) Having an estimate of the population proportion raises the minimum sample size needed. B.) Having an estimate of the population proportion reduces the minimum sample size needed. C.) Having an estimate of the population proportion has no effect on the minimum sample size needed.
Correct Answers: *Part 1 (a):* *4,148* *Part 2 (b):* *2,847* *Part 3 (c):* B.) Having an estimate of the population proportion reduces the minimum sample size needed. (*6.3.17-T*)
A cheese processing company wants to estimate the mean cholesterol content of all one-ounce servings of a type of cheese. The estimate must be within 0.76 milligram of the population mean. *Part 1 (a):* Determine the minimum sample size required to construct a 95% confidence interval for the population mean. Assume the population standard deviation is 3.15 milligrams. (Round answer *up* to the *nearest whole number* (*zero* decimal places).) *Part 2 (b):* The sample mean is 29 milligrams. Using the minimum sample size with a 95% level of confidence, does it seem likely that the population mean could be within 3% of the sample mean? within 0.3% of the sample mean? Explain. (Round all answers to *two* decimal places.) *(I didn't copy the accompanying Standard Normal Tables because they're too big.)* *Part 1 (a):* The minimum sample size required to construct a 95% confidence interval is *__* servings. *Part 2 (b):* The 95% confidence interval is (*__(1)__*, *__(2)__*). It *__(3)__* likely that the population mean could be within 3% of the sample mean because the interval formed by the values 3% away from the sample mean *______(4)______* the confidence interval. It *__________(5)__________* [seem] likely that the population mean could be within 0.3% of the sample mean because the interval formed by the values 0.3% away from the sample mean *____________(6)____________* the confidence interval.
Correct Answers: *Part 1 (a):* *66* *Part 2 (b):* *(1):* *28.24* *(2):* *29.76* *(3):* *seems* *(4):* *entirely contains* *(5):* *does not seem* *(6):* *overlaps but does not entirely contain* (*6.1.49*)
Translate the statement into a confidence interval (round answers to *three* decimal places). Approximate the level of confidence (round answer to *one* decimal place). In a survey of *1,035 adults* in a country, *71%* said being able to speak the language is at the core of national identity. The survey's margin of error is *±3.5%*. *Part 1:* The confidence interval for the proportion is (*__(1)__*, *__(2)__*). *Part 2:* The interval is a *___%* confidence interval.
Correct Answers: *Part 1:* *(1):* *0.675* *(2):* *0.745* *Part 2:* *98.7%* (*6.3.29-T*)
You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean (round all answers to *two* decimal places). Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. *A random sample of 45 home theater systems has a mean price of $124.00. Assume the population standard deviation is $17.30.* *Part 1:* Construct a 90% confidence interval for the population mean. The 90% confidence interval is (*___(1)___*, *___(2)___*). *Part 2:* Construct a 95% confidence interval for the population mean. The 95% confidence interval is (*___(1)___*, *___(2)___*). *Part 3:* Interpret the results. Choose the correct answer below. A.) With 90% confidence, it can be said that the sample mean price lies in the first interval. With 95% confidence, it can be said that the sample mean price lies in the second interval. The 95% confidence interval is wider than the 90%. B.) With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%. C.) With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is narrower than the 90%.
Correct Answers: *Part 1:* *(1):* *119.76* *(2):* *128.24* *Part 2:* *(1):* *118.95* *(2):* *129.05* *Part 3:* B.) With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%. (*6.1.35*)
In a random sample of 27 people, the mean commute time to work was 31.2 minutes and the standard deviation was 7.1 minutes. Assume the population is normally distributed and use a *t* - distribution to construct a 90% confidence interval for the population mean *μ*. What is the margin of error of *μ*? (Round both answers to *one* decimal place.) Interpret the results. *Part 1:* The confidence interval for the population mean *μ* is (*__(1)__*, *__(2)__*). *Part 2:* The margin of error of *μ* is *__*. *Part 3:* Interpret the results. A.) With 90% confidence, it can be said that the population mean commute time is between the bounds of the confidence interval. B.) It can be said that 90% of people have a commute time between the bounds of the confidence interval. C.) With 90% confidence, it can be said that the commute time is between the bounds of the confidence interval. D.) If a large sample of people are taken approximately 90% of them will have commute times between the bounds of the confidence interval.
Correct Answers: *Part 1:* *(1):* *31.9* *(2):* *36.9* *Part 2:* *2.5* *Part 3:* A.) With 90% confidence, it can be said that the population mean commute time is between the bounds of the confidence interval. (*6.2.17-T*)
Use the confidence interval to find the margin of error and the sample mean. *(0.664, 0.860)* *Part 1:* The margin of error is *____*. *Part 2:* The sample mean is *____*.
Correct Answers: *Part 1:* *0.098* *Part 2:* *0.762* (*6.1.25*)
