Chapter 7 Review Confidence Intervals MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Suppose that you wish to obtain a confidence interval for a population mean. Under the conditions described below, should you use the z-interval procedure, the t-interval procedure, or neither? 1) - The population is far from being normally distributed. - the sample size is large. A) t-interval procedure B) Neither C) z-interval procedure 2) Suppose that you wish to obtain a confidence interval for a population mean. Under the conditions described below, should you use the z-interval procedure, the t-interval procedure, or neither? 2) - The population standard deviation is unknown. - The population is normally distributed. - The sample size is small. A) z-interval procedure B) Neither C) t-interval procedure Determine the critical value zα/2 that corresponds to the given level of confidence. 3) 88% A) 1.555 B) 0.81 C) 1.21 D) 1.175 3) Solve the problem. 4) What is zα/2 when α = 0.05? A) 1.645 B) 2.575 C) 2.33 D) 1.96 4) 5) Find zα/2 for the given value of α. α = 0.14 A) 0.14 B) 1.48 C) 1.08 D) 1.58 5) Find the t-value. 6) Find the critical t-value tα/2 that corresponds to 95% confidence and n = 16. A) 2.947 B) 2.131 C) 2.602 D) 1.753 6) 7) Find the critical t-value tα/2 that corresponds to 90% confidence and n = 15. A) 2.145 B) 1.345 C) 2.624 D) 1.761 7) 1
Solve the problem. 8) A 90% confidence interval for the average salary of all CEOs in the electronics industry was constructed using the results of a random survey of 45 CEOs. The interval was ($139,048, $154,144). Give a practical interpretation of the interval. A) We are 90% confident that the mean salary of all CEOs in the electronics industry falls in the interval $139,048 to $154,144. B) 90% of all CEOs in the electronics industry have salaries that fall between $139,048 to $154,144. C) 90% of the sampled CEOs have salaries that fell in the interval $139,048 to $154,144. D) We are 90% confident that the mean salary of the sampled CEOs falls in the interval $139,048 to $154,144. 8) 9) A retired statistician was interested in determining the average cost of a $200,000.00 term life insurance policy for a 60-year-old male non-smoker. He randomly sampled 65 subjects (60-year-old male non-smokers) and constructed the following 95 percent confidence interval for the mean cost of the term life insurance: ($850.00, $1050.00). State the appropriate interpretation for this confidence interval. Note that all answers begin with "We are 95 percent confidence that " A) The average term life insurance costs for all 60-year-old male non-smokers falls between $850.00 and $1050.00 B) The average term life insurance cost for sampled 65 subjects falls between $850.00 and $1050.00 C) The term life insurance cost of the retired statistician's insurance policy falls between $850.00 and $1050.00 D) The term life insurance cost for all 60-year-old male non-smokers' insurance policies falls between $850.00 and $1050.00 9) 10) A marketing research company is estimating the average total compensation of CEOs in the service industry. Data were randomly collected from 18 CEOs and the 98% confidence interval for the mean was calculated to be ($2,181,260, $5,836,180). What additional assumption is necessary for this confidence interval to be valid? A) The distribution of the sample means is approximately normal. B) The population of total compensations of CEOs in the service industry is approximately normally distributed. C) The sample standard deviation is less than the degrees of freedom. D) None. The Central Limit Theorem applies. 10) 11) To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nicotine found in all commercial brands of cigarettes. A new cigarette has recently been marketed. The FDA tests on this cigarette gave a mean nicotine content of 28.4 milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes. The FDA claims that the mean nicotine content exceeds 31.0 milligrams for this brand of cigarette, and their stated reliability is 90%. Do you agree? A) Yes, since the value 31.0 does fall in the 90% confidence interval. B) No, since the value 31.0 does not fall in the 90% confidence interval. C) Yes, since the value 31.0 does not fall in the 90% confidence interval. D) No, since the value 31.0 does fall in the 90% confidence interval. 11) 2
Solve the problem. 12) Fifteen SmartCars were randomly selected and the highway mileage of each was noted. The analysis yielded a mean of 47 miles per gallon and a standard deviation of 5 miles per gallon. Which of the following would represent a 90% confidence interval for the average highway mileage of all SmartCars? 12) A) 47 ± 1.753 5 15 B) 47 ± 1.761 5 15 C) 47 ± 1.345 5 15 D) 47 ± 1.645 5 15 Use the given degree of confidence and sample data to construct a confidence interval for the population mean μ. Assume that the population has a normal distribution. 13) n = 10, x = 9.8, s = 2.4, 95 percent A) 8.08 < μ < 11.52 B) 8.11 < μ < 11.49 C) 8.09 < μ < 11.51 D) 8.41 < μ < 11.19 13) 14) Thirty randomly selected students took the calculus final. If the sample mean was 92 and the standard deviation was 9.4, construct a 99 percent confidence interval for the mean score of all students. A) 89.08 < μ < 94.92 B) 87.27 < μ < 96.73 C) 87.29 < μ < 96.71 D) 87.77 < μ < 96.23 14) 15) The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were: 13.2 5.1 7.5 8.0 12.7 7.6 13.8 14.5 7.7 10.5 Determine a 95 percent confidence interval for the mean time for all players. A) 12.30 < μ < 7.82 B) 7.82 < μ < 12.30 C) 7.72 < μ < 12.40 D) 12.40 < μ < 7.72 15) Find the confidence interval specified. Assume that the population is normally distributed. 16) A random sample of 30 households was selected from a particular neighborhood. The number of cars for each household is shown below. Estimate the mean number of cars per household for the population of households in this neighborhood. Give the 95% confidence interval. 16) 2 0 1 2 3 2 1 0 1 4 1 3 2 0 1 1 2 3 1 2 1 0 0 3 2 2 2 1 0 2 A) 1 to 2 cars B) 1.5 to 1.9 cars C) 1.1 to 1.9 cars D) 1.3 to 1.7 cars SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Solve the problem. 17) To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nicotine found in all commercial brands of cigarettes. A new cigarette has recently been marketed. The FDA tests on this cigarette yielded a mean nicotine content of 24.5 milligrams and standard deviation of 2.3 milligrams for a sample of n = 82 cigarettes. Find a 95% confidence interval for μ. 17) 18) Construct a 98% confidence interval for the population mean, μ. Assume the population has a normal distribution. A study of 14 car owners showed that their average repair bill was $192 with a standard deviation of $8. Round to the nearest cent. 18) 3
19) A local bank needs information concerning the savings account balances of its customers. A random sample of 15 accounts was checked. The mean balance was $686.75 with a standard deviation of $256.20. Find a 98% confidence interval for the true mean. Assume that the account balances are normally distributed. Round to the nearest cent. 19) MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Solve the problem. 20) A marketing research company is estimating which of two soft drinks college students prefer. A random sample of 157 college students produced the following confidence interval for the proportion of college students who prefer drink A: (.344,.494). Is this a large enough sample for this analysis to work? A) Yes, since n = 157 (which is 30 or more). B) No. C) Yes, since both np^ 5 and n(1-p^) 5 D) It is impossible to say with the given information. 21) What type of car is more popular among college students, American or foreign? One hundred fifty-nine college students were randomly sampled and each was asked which type of car he or she prefers. A computer package was used to generate the printout below for the proportion of college students who prefer American automobiles. 20) 21) SAMPLE PROPORTION =.396226 SAMPLE SIZE = 159 Is the sample large enough for the interval to be valid? A) Yes, since both np^ 5 and n(1-p^) 5 B) No, the sample size should be at 10% of the population. C) No, the population of college students is not normally distributed. D) Yes, since n > 30. 22) What type of car is more popular among college students, American or foreign? One hundred fifty-nine college students were randomly sampled and each was asked which type of car he or she prefers. A computer package was used to generate the printout below of a 90% confidence interval for the proportion of college students who prefer American automobiles. 22) SAMPLE PROPORTION =.396 SAMPLE SIZE = 159 UPPER LIMIT =.460 LOWER LIMIT =.332 Based on the interval above, do you believe that 21% of all college students prefer American automobiles? A) No, and we are 90% confident of it. B) Yes, and we are 100 %sure of it. C) No, and we are 100% sure of it. D) Yes, and we are 90% confident of it. 4
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. 23) A newspaper reports on the topics that teenagers most want to discuss with their parents. The findings, the results of a poll, showed that 46% would like more discussion about the family's financial situation, 37% would like to talk about school, and 30% would like to talk about religion. These and other percentages were based on a national sampling of 505 teenagers. Estimate the proportion of all teenagers who want more family discussions about religion. Use a 99% confidence level. 23) 24) A random sample of 50 employees of a large company was asked the question, "Do you participate in the company's stock purchase plan?" The answers are shown below. 24) yes no no yes no no yes yes no no no yes yes yes no yes no no yes yes no yes yes no yes yes no yes yes yes yes no no yes yes yes yes yes no yes no yes yes no yes yes yes yes yes yes Use a 90% confidence interval to estimate the proportion of employees who participate in the company's stock purchase plan. 25) A survey of 700 non-fatal accidents showed that 163 involved uninsured drivers. Construct a 99% confidence interval for the proportion of fatal accidents that involved uninsured drivers. Round to the nearest thousandth. 25) 26) In a survey of 10 musicians, 2 were found to be left-handed. Is it practical to construct the 90% confidence interval for the population proportion, p? Explain. 26) 27) A survey of 300 fatal accidents showed that 123 were alcohol related. Construct a 98% confidence interval for the proportion of fatal accidents that were alcohol related. 27) MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 28) A survey of 2450 golfers showed that 281 of them are left-handed. Construct a 98% confidence interval for the proportion of golfers that are left-handed. A) (0.683, 0.712) B) (0.369, 0.451) C) (0.203, 0.293) D) (0.100, 0.130) 28) 5
Answer Key Testname: CH 7 CONFIDENCE INTERVALS REVIEW 1) C 2) C 3) A 4) D 5) B 6) D 7) B 8) A 9) A 10) B 11) B 12) B 13) A 14) B 15) C 16) C 17) For confidence coefficient.95, 1 - α =.95 α = 1 -.95 =.05. α/2 =.05/2 =.025. zα/2 = z.025 = 1.96. The 95% confidence interval is: x ± zα/2 s n = 24.5 ± 1.96 2.3 82 24.5 ±.498 = (24.002, 24.998) 18) ($186.33, $197.67) 19) ($513.17, $860.33) 20) C 21) A 22) A 23) For confidence coefficient.99, 1 - α =.99 α = 1 -.99 =.01. α/2 =.01/2 =.005. zα/2 = z.005 = 2.575. The 99% confidence interval for p is: p^ ± zα/2 p^q^ n.30 ± 2.575.30(.70) 505.30 ±.0525 24) p^ = 32 50 =.64; The confidence interval is.64 ± 1.645 (.64)(.36).64 ±.112. 50 25) (0.192, 0.274) 26) It is not practical to find the confidence interval. It is necessary that np^(1 - p^) 10 to insure that the distribution of p^ be normal. (np^(1 - p^) = 1.6) 27) (0.344, 0.476) 28) D 6