Question.3179 - 10.7 According to a recent study, when shopping online for luxury goods, men spend a mean of $2,401, whereas women spend a mean of $1,527 (data extracted from R. A. Smith, “Fashion Online: Retailers Tackle the Gender Gap,” The wall Street Journal, March 13, 2008, pp. D1, D10). Suppose that the study was based on a sample of 600 men and 700 females, and the standard deviation of the amount spent was $1,200 for men and $1,000 for women. a. State the null and alternative hypothesis if you want to determine whether the mean amount spent is higher for men than for women. b. In the context of this study, what is the meaning of the Type I error? c. In the context of this study, what is the meaning of the Type II error? d. At the 0.01 level of significance, is there evidence that the mean amount spent is higher for men than for women? 10.11 Digital cameras have taken over the majority of the point-and-shoot camera market. One of the important features of a camera is the battery life, as measured by the number of shots taken unit the battery needs to be recharged. The file DigitalCameras contains the battery life of 29 sub-compact cameras and 16 compact cameras (data extracted from “Digital Cameras,” Consumer Reports, July 2009, pp. 28-29). a. Assuming that the population variances from both types of digital cameras are equal is there evidence of a difference in the mean battery life between the two types of digital cameras ()? b. Determine the p-value in (a) and interpret its meaning. c. Assuming that the population variances from both types of digital cameras are equal, construct and interpret a 95% confidence interval estimate of the difference between the population mean battery life of the two types of digital cameras. 10.45 A professor in the accounting department of a business school claims that there is more variability in the final exam scores of students taking the introductory accounting course who are not majoring in accounting than for students taking the course who are majoring in accounting. Random samples of 13 non-accounting majors and 10 accounting majors are selected from the professor’s class roster in his large lecture, and the following results are computed based on the final exam scores: Non-accounting: = 13 = 210.2 Accounting: = 10 = 36.5 a. At the 0.05 level of significance, is there evidence to support the professor’s claim? b. Interpret the p-value. c. What assumption do you need to make in (a) about the two populations in order to justify your use of the F test? 10.47 A bank with a branch located in a commercial district of a city has developed an improved process for serving customers during the noon-to-1 pm. lunch period. The waiting time (defined as the time elapsed form when the customer enters the line until he or she reaches the teller window) needs to be shortened to increase customer satisfaction. A random sample of 15 customers is selected (and stored in Bank 1), and the results (in minutes) are as follows: 4.21 5.55 3.02 5.13 4.77 2.34 3.54 3.20 4.50 6.10 0.38 5.12 6.46 6.19 3.79 Suppose that another branch, located in a residential area, is also concerned with the noon-to-1 pm. lunch period. A random sample of 15 customers is selected (and stored in Bank2), and the results (in minutes) are as follows: 9.66 5.90 8.02 5.79 8.73 3.82 8.01 8.35 10.49 6.68 5.64 4.08 6.17 9.91 5.47 a. Is there evidence of a difference in the variability of the waiting time between the two branches? (Use () b. Determine the p-value in (a) and interpret its meaning. c. What assumption about the population distribution of the two banks is necessary in (a)? is the assumption valid for these data? d. Based on the results of (a), is it appropriate to use the pooled-variance t test to compare the means of the two branches?
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10.7 Sol: a) The null and alternative hypothesis to determine whether the mean amount spend is higher for men than for women: null hypothesis, H 0 : m 1 <m 2 alternative hypothesis, H A :m 1 >m 2 where m 1 = mean amount spend by men and m 2 = mean amount spend by women b) Type 1 error with context to the problem would be rejecting the null hypothesis when mean amount spend by men is lower than for women. c) Type 2 error with context to the problem would be accepting the null hypothesis when mean amount spend by men is higher than for women. d) We need to use the statistic: Z=( m 1 -m 2 )/√(s 1 2 /n 1 +s 2 2 /n 2 ) Where, m 1 =$2401 m 2 =$1527 s 1 =$1200 s 2 =$1000 n 1 =600 n 2 =700 putting all the values we get, Z=14.125 based on 0.01 significance level, the critical region lies between (-2.33,+2.33). The value 14.125 does not lie between the critical region Hence we reject the null hypothesis. Therefore, the mean amount spend by men is higher than for women. 10.11 Sol: a) Batter y Life Camera Type Subcompac t Compac t 320 Subcompac t 320 520 520 Subcompac t 520 260 160 Subcompac t 160 400 160 Subcompac t 160 200 300 Subcompac t 300 300 120 Subcompac t 120 150 520 Subcompac 520 360 t 440 Subcompac t 440 200 300 Subcompac t 300 260 170 Subcompac t 170 80 150 Subcompac t 150 200 300 Subcompac t 300 260 180 Subcompac t 180 400 100 Subcompac t 100 260 150 Subcompac t 150 200 400 Subcompac t 400 170 Subcompac t 170 180 Subcompac t 180 160 Subcompac t 160 240 Subcompac t 240 320 Subcompac t 320 260 Subcompac t 260 150 Subcompac t 150 320 Subcompac t 320 180 Subcompac t 180 140 Subcompac t 140 220 Subcompac t 220 SUBCOMPACT COMPACT Mean 245.862(m 1 ) 258.125(m 2 ) Variance 12989.79 12103.5 Standard deviation, s=√((n 1 s 1 2 +n 2 s 2 2 )/(n 1 +n 2 -2)) Where, s 1 = 13453.71, s 2 =12910.4, n 1 =29, n 2 =16 S= 13568.0215 Now the statistic, t= (m 1 -m 2 )/s √(1/n 1 +1/n 2 ) = -0.08768 t critical at α=0.05 =2.02 > t Therefore, there is insufficient evidence to reject the null hypothesis. b) The p value is 0.415. That means there is a 41.5% chance that any difference seen here is due to sampling error. 10.45 a) Ho: σ²1 ≤ σ²2 Ha: σ²1 > σ²2 F-statistic = 210.2/36.5 = 5.7589 Numerator d.f. = 12, denominator d.f. = 9 Critical value of F = 3.0729 Since F-statistic lies in critical region, null hypothesis is rejected. Non-accounting variance is greater than that of accounting, and professor claim is supported by the evidence. b) P-value = 0.0066 This means that if first variance was actually not greater than the second, the probability that observed variances would have occurred is 0.0066. c) Assumption of normality and independence of samples 10.47 a) Ho: µ1 = µ2 Ha: µ1 ≠ µ2 Mean1 = 4.2867, Mean2 = 7.1147, s1 = 1.638, s2 = 2.0822 t-statistic = (4.2867-7.1147)/sqrt((1.638^2+2.0822^2)/15) = -4.1343 Degree of freedom = 15+15-2 = 28 Critical values = ±2.0484 Since test statistic lies in critical region, null hypothesis is rejected. There is a difference in mean waiting times. b) P-value = 0.0003 This means that if actually there was no significant difference in waiting times, then the probability that the observed difference would have occurred was 0.0003. c) Two populations are normally distributed and samples are independent and random. d) For 95% confidence, with d.f. of 28, critical t = 2.0484 Lower limit = (4.2867-7.1147) - 2.0484 * sqrt((1.638^2+2.0822^2)/15) = -4.22918 Upper limit = (4.2867-7.1147) + 2.0484 * sqrt((1.638^2+2.0822^2)/15) = -1.42682 So, 95% confidence interval for difference of mean is (-4.22918, 1.42682).More Articles From Statistics
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