e 95 Confidence Interval for \u03bc 175 21312516 175 1332 16168 18832 Problem 2 6

E 95 confidence interval for μ 175 21312516 175 1332

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(e)95% Confidence Interval for μ: 175 ± (2.131)(25/√16) = 175 ± 13.32 = (161.68, 188.32).Problem 26
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From all the members of a club who played the same game, we randomly chose 8 males and 8 females and recorded their scores as follows:Male: 77 74 82 73 87 69 66 80(sample variance = 48.0)Female: 72 68 76 68 84 68 61 76(sample variance = 49.1)Assume that the distributions of the scores of the male members and the female members are independently normally distributed with the same variance. We want to test whether the average score of the male members is different from that of the female members in the club.(a)State the appropriate null and alternative hypotheses. (b)What is the point estimate of the difference between the average scores of the male and female members? (c)Compute the value of the test statistic. (d)Develop a 99% confidence interval for the difference between the average scores of the male and female members. (e)Does there exist a significant difference between the average scores of the male and female members at 1% level of significance? Explain. a. H0: μ1-μ2= 0Ha: μ1-μ20b. = 76 – 71.625 = 4.375c. d. e. Since “0” is within the 99% confidence interval, do not reject H0. Conclude that the average scores of male and female members are not different at 1% level of significance.Problem 3A company provides customers opportunities to buy its products over the Internet, and after 10 months of operation, the company reported that 44% of orders were from the repeat customers. Assume that the company will use a sample of customers orders each quarter to determine whether the proportion of orders from repeat customers changed from the initial 44%.7
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(a)Formulate the null and alternative hypotheses.(b) During the first quarter a sample of 500 orders showed 205 repeat customers. What is the p-value? Using α=5%, what is your conclusion?(c)If you use α=1%, what is the conclusion?a. H0: p = 0.44, Ha: p ≠ 0.44b. = 205/500= 0.41z = (0.41-0.44)/sqrt{p0*(1-p0)/n}=-1.35 p-value = 2*(0.5-0.4115)= 0.177Since 0.177 > 0.05, we do not reject the null hypothesis.c. Since 0.177>0.01, we do not reject the null hypothesis.8
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  • Spring '10
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  • Statistics, Null hypothesis, Statistical hypothesis testing

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