1. the Mean Weight of 18-Year-Old Females Is 126 Pounds, and the Standard Deviation Is

Chapter 6:

1. The mean weight of 18-year-old females is 126 pounds, and the standard deviation is 15.7 pounds. If a sample of 25 females is selected, find the probability that the mean of the sample will be greater than 128.3 pounds. Assume the weight is normally distributed. 2. The average labor cost for car repairs for a large chain of car repair shops is $48.25. The standard deviation is $4.20. Assume the cost is normally distributed. (a) If a store is randomly selected, find the probability that the labor cost will be between $46 and $48. (b) If 36 stores are randomly selected, find the probability that the mean repair cost will be between $46 and $48. (c) Which is larger? Explain why? 3. The mean grade point average of the engineering majors at a large university is 3.23, with a standard deviation of 0.72. In a class of 48 students, (a) find the probability that the mean grade point average of the students is less than 3. (b) find the probability that the mean grade point average of the students is between 3.12 and 3.3. 4. The average time it takes a group of adults to complete a certain achievement test is 46.2 minutes. The standard deviation is 8 minutes. Assume the time is normally distributed. (a) Find the probability that a randomly selected adult will complete the test in less than 43 minutes.

(b) Find the probability that if 50 randomly selected adults take the test, the mean time it takes the group to complete the test will be less than 43 minutes. 5. A contractor states that 90% of all homes sold have burglar alarm systems. If a contractor sells 250 homes, find the probability that at least 210 of them will have burglar alarm system. [Use Normal approximation.] 6. A political candidate estimates that 30% of the voters in his party favor his proposed tax reform bill. If there are 400 people at a rally, find the probability that at least 100 favor his tax bill. [Use Normal approximation.]

Answer: 1. 0.2327 2. 0.1815, 0.3598; The distribution of the sample means is much less variable than the distribution of individual value.

3. a. 0.0134, b. 0.6048; 4. a. 0.3446, b. 0.0023; 5. 0.9995; 6. 0.9873. 2

Chapter 7:

1. The inspection division of the Lee County Weights and Measures Department is interested in estimating the actual amount of soft drink that is placed in 2-liter bottles at the local bottling plant of a large nationally known soft drink company. The bottling plant has informed the inspection division that the standard deviation for 2- liter bottles is 0.05 liter. A random sample on 100 2-liter bottles obtained from this bottle plant indicated a sample average of 1.99 liters.

(a) Find a 95% confidence interval for the true average amount of soft drink in each bottle. (b) Find the maximum error. (c) If we want the maximum error to be 0.0075 liter with 95% confidence, how large the sample size should be? 2. The personnel department of a large corporation would like to estimate the family dental expenses of its employees in order to determine the feasibility of providing a dental insurance plan. A random sample of 10 employees revealed the following family dental expenses in the preceding year:

$110, 362, 246, 85, 510, 208, 173, 425, 316, 179.

Find a 90% confidence interval for the true mean dental expenses for all employees of this corporation. Assume that the dental expenses is normally distributed.

3. An advertising agency that serves a major radio station would like to estimate the average amount of time that the station’s audience spends listening to radio on a daily basis. What sample size is needed if the agency wants to be 90% confident of being correct to within 5 minutes? From past studies, the standard deviation is about 45 minutes. 4. The real estate assessor for a county government wishes to study various characteristics concerning single-family houses in the county. A random sample of 70 houses revealed the following:

Heating area of the house : average = 1,759 sq ft, s = 380 ft. 42 houses had central air conditioning.

i. Find a 99% confidence interval for the true average heating area of the house. ii. Find a 95% confidence interval for the proportion of houses that have central air conditioning. 3

5. A researcher wishes to be 92% confident that her estimate of the true proportion of individuals who travel overseas is within 0.04 of the true proportion. (a) In a prior study, a sample of 200 people showed that 80 traveled overseas last year. Find the required sample size. (b) If no estimate of the sample proportion is available, Find the required sample size.

6. A random sample of stock prices per share is shown. Find a 95% confidence interval for the variance and standard deviation for the prices. Assume the price is normally distributed.

$ 26.69 $13.88 $28.37 $12.00 75.37 7.50 47.50 43.00 40.25 10.87 46.12 14.75

7. A sample of 18 automobiles has a pollution by-product release standard deviation of 2.3 ounces when one gallon of gasoline is used. Find the 90% confidence interval of the population standard deviation.

Answer:

1. a (1.9802, 1.99980) b. 0.0098, c. 171;

2. ($180.94, $341.86); 3. 220; 4. a. (1642 sq ft, 1876 sq ft), b. (0.485, 0.715) or (48.5%, 71.5%);

5. a. 460, b. 479;

6. For variance ( 209.66, 1204.33), for standard deviation ($14.48, $34.7);

7. (1.81 ounces, 3.22 ounces); 4

Chapter 8:

1.A researcher claims that the average cost of men’s athletic shoes is less than $80. He selects a random sample of 36 pairs of shoes from a catalog and finds the mean is $75 and the standard deviation is $19.2. Is there enough evidence to support the researcher’s claim at α = 0.10? i. Explain α = 0.10 in words. ii. State the null and the alternative hypothesis in symbols and in words. iii. Find the p-value. iv. Make the decision. v. State your conclusion in words.

2. A statistician claims that the average age of people who purchases lottery tickets is 70. A random sample of 30 is selected, and their ages are recorded. At α = 0.05 is there enough evidence to reject the statistician’s claim?

49 90 80 24 61 79 68 63 72 46 65 76 71 56 70 71 71 67 52 82 74 39 49 69 22 56 70 74 62 45

(a) Use TI-83/84 to create a data list. (b) Find the mean and the standard deviation. (c) State the null and the alternative hypotheses in symbols and in words. (d) Find the p-value from TI-83/84 and explain it in words. (e) Use the p-value to make a decision. (f) State your conclusion in words.

3.A recent survey stated that the average single-person household received at least 37 telephone calls per month. To test the claim, a researcher surveyed 29 single-person households and found that the average number of calls was 34.9. The standard deviation of the sample was 6. At α = 0.05, can the claim be rejected?

a. State the null and alternative hypotheses in symbols and in words. b. Use TI-83/84 to find the p-value. c. Make a decision. d. State your conclusion in words.

4.A recent study found that at least 15% of all eighth-grade students are overweight. In a sample of 80 students, 9 were found to be overweight. a. Find a 90% confidence interval for the proportion of all overweight eighth-grade students. 5

b. At 5% significance level, can we conclude that at least 15% of all eighth- grade students are overweight? State your null and alternative hypotheses, summarize information from the problem, use TI-83/84 to find the p-value, and present your conclusion in words.

Answers:

1. a. 10% of the time, we allow ourselves to reject the null hypothesis when it is actually true. b. Null hypothesis μ  $80 (The claim is false.) vs. Alternative μ < $80 (The claim is true.). c p-value = .0590 d Reject the null hypothesis since p-value = .0590 < α = 0.10. e At 10% level of significance, we have enough evidence to support the researcher’s claim that the average cost of men’s athletic shoes is less than $80.

2. b. mean = 62.43, s.d. = 16.07, c. Null Hypothesis: μ = 70 (The average age is 70.) Alternative Hypothesis :   70 (The average age is not 70.) d. p-value = 0.0099, e. Since p-value=0.0099 < 0.05 = α, we reject the null hypothesis. f. At 5% level of significance, we conclude that the researcher’s claim is false. That is, the average age of people who purchases lottery tickets is not 70.

3. a. The Null hypothesis: μ  37 (The average single-person household receives at least 37 calls.) The Alternative hypothesis : μ < 37 (The average single-person household receives less than 37 calls.) b. p-value = 0.0349 c. Reject the null hypothesis, since p-value = 0.0349 < α = 0.05. d. At 5% level of significance, we conclude that the researcher’s claim is false. That is, the average single-person household received less than 37 calls per month.

4. (a) We are 90% confident that the proportion of all overweight eighth grade students is between 5.4% and 17%. (b) Null hypothesis: P ≥ 0.15 Alternative hypothesis : P < 0.15 p-value = 0.1738 > 0.05 = α, do not reject the null hypothesis. At 5% level of significance, we do not have enough evidence to conclude that less than 15% of eighth grade students is overweight.