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6. On average, 30-minute television sitcoms have 22 minutes of programming (CNBC, February 23, 2006). Assume that the probability distribution for minutes of programming can be approximated by a uniform distribution from 18 minutes to 26 minutes. a. What is the probability a sitcom will have 25 or more minutes of programming? b. What is the probability a sitcom will have between 21 and 25 minutes of programming? c. What is the probability a sitcom will have more than 10 minutes of commercials or other nonprogramming interruptions? 12. Given that z is a standard normal random variable, compute the following probabilities. a. P(0 ≤ z ≤ .83) b. P(-1.57≤ z≤ 0) c. P(z > .44) Page 2 of 14 d. P(z ≥ -.23) e. P(z < 1.20) f. P(z ≤ -.71) 14. Given that z is a standard normal random variable, find z for each situation. a. The area to the left of z is .9750. b. The area between 0 and z is .4750. c. The area to the left of z is .7291. d. The area to the right of z is .1314. e. The area to the left of z is .6700. Page 3 of 14 f. The area to the right of z is .3300. 17. For borrowers with good credit scores, the mean debt for revolving and installment accounts is $15,015 (BusinessWeek, March 20, 2006). Assume the standard deviation is $3540 and that debt amounts are normally distributed. a. What is the probability that the debt for a borrower with good credit is more than $18,000? b. What is the probability that the debt for a borrower with good credit is less than$10,000? c. What is the probability that the debt for a borrower with good credit is between $12,000 and $18,000? d. What is the probability that the debt for a borrower with good credit is no more than $14,000? Page 4 of 14 Homework 6 13. A sample of 5 months of sales data provided the following information: Month: 1 2 3 4 5 Units Sold: 94 100 85 94 92 a. Develop a point estimate of the population mean number of units sold per month. n = x̄= Point estimate of the population mean = b. Develop a point estimate of the population standard deviation Point estimate of the population standard deviation = ____ xi (xi- x̄)² 17. The American Association of Individual Investors (AAII) polls its subscribers on a weekly basis to determine the number who are bullish, bearish, or neutral on the short-term prospects for the stock market. Their findings for the week ending March 2, 2006, are consistent with the following sample results (AAII website, March 7, 2006). Bullish 409 Neutral 299 Bearish 291 Develop a point estimate of the following population parameters. Page 5 of 14 a. The proportion of all AAII subscribers who are bullish on the stock market. b. The proportion of all AAII subscribers who are neutral on the stock market. c. The proportion of all AAII subscribers who are bearish on the stock market. 18. A population has a mean of 200 and a standard deviation of 50.Asample of size 100 will be taken and the sample mean will be used to estimate the population mean. a. What is the expected value of x̄ ? b. What is the standard deviation of x̄ ? c. Show the sampling distribution of x̄. d. What does the sampling distribution of x̄show? 31. A sample of size 100 is selected from a population with p = .40. a. What is the expected value of 𝑝𝑝̅? Page 6 of 14 b. What is the standard error of 𝑝𝑝̅? c. Show the sampling distribution of 𝑝𝑝̅. d. What does the sampling distribution of 𝑝𝑝̅show? Homework 7 7. The Wall Street Journal reported that automobile crashes cost the United States $162billion annually (The Wall Street Journal, March 5, 2008). The average cost per person for crashes in the Tampa, Florida, area was reported to be $1599. Suppose this average cost was based on a sample of 50 persons who had been involved in car crashes and that the population standard deviation is σ = $600. What is the margin of error for a 95% confidence interval? What would you recommend if the study required a margin of error of $150 or less? 15. Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 65 weekly reports showed a sample mean of 19.5 customer contacts per Page 7 of 14 week. The sample standard deviation was 5.2. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel. 90% confidence interval: 95% confidence interval: 36. According to statistics reported on CNBC, a surprising number of motor vehicles are not covered by insurance (CNBC, February 23, 2006). Sample results, consistent with the CNBC reports, showed 46 of 200 vehicles were not covered by insurance. a. What is the point estimate of the proportion of vehicles not covered by insurance? b. Develop a 95% confidence interval for the population proportion. Page 8 of 14 Chapter 9 6. The label on a 3-quart container of orange juice states that the orange juice contains an average of 1 gram of fat or less. Answer the following questions for a hypothesis test that could be used to test the claim on the label. a. Develop the appropriate null and alternative hypotheses. b. What is the Type I error in this situation? What are the consequences of making this error? c. What is the Type II error in this situation? What are the consequences of making this error? 10. Consider the following hypothesis test: A sample of 40 provided a sample mean of 26.4. The population standard deviation is 6. a. Compute the value of the test statistic. b. What is the p-value? Page 9 of 14 c. At α = .01, what is your conclusion? Homework 8 2. Consider the following hypothesis test. H0: μ1 - μ2 ≤ 0 Ha: μ1 - μ2 > 0 The following results are for two independent samples taken from the two populations. a. What is the value of the test statistic? b. What is the p-value? c. With α = .05, what is your hypothesis testing conclusion? Page 10 of 14 4. Condé Nast Traveler conducts an annual survey in which readers rate their favorite cruise ship. All ships are rated on a 100-point scale, with higher values indicating better service. A sample of 37 ships that carry fewer than 500 passengers resulted in an average rating of 85.36, and a sample of 44 ships that carry 500 or more passengers provided an average rating of 81.40 (Condé Nast Traveler, February 2008). Assume that the population standard deviation is 4.55 for ships that carry fewer than 500 passengers and 3.97 for ships that carry 500 or more passengers. a. What is the point estimate of the difference between the population mean rating for ships that carry fewer than 500 passengers and the population mean rating for ships that carry 500 or more passengers? b. At 95% confidence, what is the margin of error? c. What is a 95% confidence interval estimate of the difference between the population mean ratings for the two sizes of ships? Page 11 of 14 6. Are nursing salaries in Tampa, Florida, lower than those in Dallas, Texas? Salary data show staff nurses in Tampa earn less than staff nurses in Dallas (The Tampa Tribune, January 15, 2007). Suppose that in a follow-up study of 40 staff nurses in Tampa and 50 staff nurses in Dallas you obtain the following results. a. Formulate hypothesis so that, if the null hypothesis is rejected, we can conclude that salaries for staff nurses in Tampa are significantly lower than for those in Dallas. Use α = .05. b. What is the value of the test statistic? c. What is the p-value? d. What is your conclusion? Page 12 of 14 Homework 9 4. M&M/MARS, maker of M&M chocolate candies, conducted a national poll in which more than 10 million people indicated their preference for a new color. The tally of this poll resulted in the replacement of tan-colored M&Ms with a new blue color. In the brochure “Colors” made available by M&M/MARS Consumer Affairs, the distribution of colors for the plains candies is as follows: Brown Yellow Red Orange Green Blue 30% 20% 20% 10% 10% 10% In a follow-up study, samples of 1-pound bags were used to determine whether the reported percentages were indeed valid. The following results were obtained for one sample of 506 candies. Brown Yellow Red Orange Green Blue 177 135 79 41 36 38 Use α = .05 to determine whether these data support the percentages reported by the company. H0:___________ Ha:___________ Category Observed Frequency (fi) Expected Frequency (ei) (fi – ei) 2 /ei Brown Yellow Red Orange Green Blue Totals Chi-Square = p-value = Conclusion: 4. A Bloomberg Business week subscriber study asked, “In the past 12 months, when traveling for business, what type of airline ticket did you purchase most often?” Sample data obtained are shown in the following contingency table. Page 13 of 14 Using α = .05 level of significance, is the type of ticket purchased independent of the type of flight? What is your conclusion? H0:____________________ Ha:____________________ Observed Frequency (fi) Category Domestic International Total First Business Full Fare Total Expected Frequency (ei) Category Domestic International Total First Business Full Fare Total Chi-Square = p-value = Conclusion: 3. Given are five observations collected in a regression study on two variables. xi 2 6 9 13 20 yi 7 18 9 26 23 Page 14 of 14 a. Develop a scatter diagram for these data. b. Develop the estimated regression equation for these data. c. Use the estimated regression equation to predict the value of y when x = 6.

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