第九章作业答案

a. If a random sample of 36 customer transactions indicates that the sample mean withdrawal amount is $172, withdrawal amount is no longer $160? (Use a 0.05 level of significance.)

b. Compute the p-value and interpret its meaning.

a. 临界值法：H0:160 H1:160（双侧检验） Z=X-172-1603036n2.4(使用Z统计量是因为1.对总体均值进行检验 2.总体标准差已知) Zc1.96 Since Z=2.4>1.96, we reject the null hypothesis and can conclude that it is significant that the population mean withdraw amount is no longer $160.b. p值法：H0:160 H1:160 p-value=2P(X172)=2P(X-n3036 2P(Z2.4)2(10.9918)0.01 (P-value is the probability of getting a sample with the mean of $172 or larger is 0.01 given H0 is true.) =0.05 Since p-value=0.010.05, we reject the null hypothesis and can conclude that it is significant that the population mean withdraw amount is no longer $160.2 The policy of a particular bank branch is that is ATMs must be stocked with enough cash to satisfy customers making withdrawals over an entire weekend. Customer goodwill depends on such services meeting customer needs. At this branch the population mean amount of money withdrawn from ATMs per customer transaction over the weekend is $160 with a population standard deviation of $30. Suppose that a random sample of 36 customer transactions is examined, and you find that the sample mean withdrawal amount is $172.

a. At the 0.05 level of significance, using the critical value approach to hypothesis testing, is there evidence to believe that the population mean withdrawal amount is greater than $160?

b. At the 0.05 level of significance, using the p-value approach to hypothesis testing, is there evidence to believe that the population mean withdrawal amount is greater

172-160)than

a. 临界值法：H0:160 H1:160（右单侧）X-172-1603036$160?

Z=n Zc1.652.4

Since Z=2.4>1.65, we reject the null hypothesis and can conclude that it is significant that the population mean withdraw amount is greater than $160.b. p值法：H0:160 H1:160 p-value=P(X172)=P(X-n3036 P(Z2.4)10.99180.0082 (P-value is the probability of getting a sample with the mean of $172 or larger is 0.0082 given H0 is true.) =0.05 Since p-value=0.00820.05, we reject the null hypothesis and can conclude that it is significant that the population mean withdraw amount is greater than $160.172-160)

3 The following data represent the monthly service fee in dollars if a customer’s account falls below the minimum required $1,500 balance for a sample of 26 banks for direct-deposit customers. BANKCOST2

12 8 5 5 6 6 10 10 9 7 10 7 7 5 0 10 6 9 12 0 5 10 8 5 5 9

a. Construct a 95% confidence interval for the population mean monthly service fee in dollars if a customer’s account falls below the minimum required balance. a. X=7.15， S=3.06, n=26s Xt(用该公式是因为1.对总体均值进行估计 2.总体标准差未知)n3.06 7.152.05957.150.6026 6.557.75 We are 95% sure that the population mean monthly service fee is somewhere between $6.55 and $7.75.

4 A manufacturer of flashlight batteries took a sample of 13 batteries BATTERIES from a day’s production and used them continuously until they failed to work. The life of the batteries in hours until failure is:

342 46 317 5 2 451 1049 631 512 266 492 562 298

a. At the 0.05 level of significance, is there evidence that the mean life of the batteries

is more than 400 hours?

b. Determine the p-value in (a) and interpret its meaning.

a. 临界值法： X=473.46, s=210.77H0:400 H1:400（右单侧） t=X-sn473.46-400210.77131.26

（用t统计量是因为：1.对均值进行检验 2.总体标准差未知) tc1.7823 Since t=1.26<1.7823, we do not reject the null hypothesis and can conclude that there is no evidence that the mean life of the batteries is more than 400 hours.b. p值法：H0:400 H1:400 p-value=P(X473.46)=P( P(t1.26) 10%0.05, we do not reject the null hypothesis and can conclude that it is not significant that the mean life of the batteries is more than 400 hours.

5 According to the Center for Work-Life Policy, a survey of 500 highly educated women who left careers for family reasons found 66% wanted to return to work (Anne Marie Chaker and Hilary Stout, “After Years off, Women Struggle to Revive Careers,” The Wall Street Journal, May 6, 2004, A1)

a. Construct a 95% confidence interval for the population proportion of highly educated women who left careers for family reasons who want to return to work. a. n=500, ps0.66

X-sn473.46-400210.7713) pszps(1ps)(由于此处是对总体比例进行区间估计，所以用该公式)n0.66(10.66) 0.660.04500 0.620.70 We are 95% sure that the population proportion of highly educated women who left careers who want to return to work is between 62% and 70%.

6 An article in The Wall Street Journal implies that more than half of all Americans 0.661.96would prefer being given $100 rather than a day off from work. This statement is based on a survey conducted by American Express Incentive Services, in which 593 of 1040 respondents indicated that they would rather have the $100 (Carlos Tejada, “Work Week,” The Wall Street Journal, July 25, 2000, A1).

a. At the 0.05 level of significance, is there evidence based on the survey data that more than half of all Americans would rather have $100 than a day off from work? b. Compute the p-value and interpret its meaning.

a. 临界值法：593 ps0.57021040H0:0.5 H1:0.5(右单侧,检验总体比例) Z=ps-ps(1ps)n0.5702-0.50.5702(10.5702)10404.53 （选用Z统计量是因为此处的变量是样本比例） Zc1.65 Since Z=4.53>1.65, we reject the null hypothesis and conclude that it is significant that more than half of all Americans would rather have $100 than a day off from work.

b. p值法：H0:0.5 H1:0.5 p-value=P(ps0.5702)=P( P(Z4.53)0 (The probability of getting a sample proportion equal to or more than 0.5702 is 0, given H0 is true.) =0.05 Since p-value=00.05, we reject the null hypothesis and conclude that it is significant that more than half of all Americans would rather have $100 than a day off from work.

1. A tire manufacturer wishes to investigate the tread life of its tires. A sample of 10 tires driven 50,000 miles revealed a sample mean of 0.32 inch of tread remaining with a standard deviation of 0.09 inch. Construct a 95 percent confidence interval for the population mean. Would it be reasonable for the manufacturer to conclude that after 50,000 miles the population mean amount of tread remaining is 0.3 inches?

ps-ps(1ps)n0.5702-0.50.5702(10.5702)1040)n10The endpoints of the confidence interval are 0.256 and 0.384. Because the value of 0.30 is in this interval, it is possible that the mean of the population is 0.30.

2. A student in public administration wants to determine the true proportion of cities that have private refuse collectors. The student wants the estimate to be within 0.10 of the population proportion, the desired level of confidence is 90 percent, and no estimate is available for the population proportion. What is the required sample size?

psZEZXts0.322.2620.090.320.0(1)n(1)n0.1022Z1.65n(1)(0.5)(10.5)68.0625E0.10The students needs a random sample of 69 cities.

3. The mean length of a small counterbalance bar is 43 millimeters. The production supervisor is concerned that the adjustments of the machine producing the bars have changed. He asked the Engineering Department to investigate. Engineering selects a random sample of 12 bars and measures each. The results are reported below in millimeters. 42 39 42 45 43 40 39 41 40 42 43 42 Is it reasonable to conclude that there has been a change in the mean length of the bars? Use the 0.02 significance level. H0:43H1:43 t=x-s/n1.78/12 tc2.71841.5-43.02.92

The null hypothesis that the population mean is 43 millimeters is rejected because the computed t of -2.92 lies in the area to the left of -2.718.