Peter Wagogo

Student’s Name Professor’s Name Course Date Stat 2606 E: Assignment 4 Question 1 n = 25 µ =10 H0: true mean net weight is equal to the net weight of 10 ounces indicated on the container H1: true mean net weight is lower than the net weight of 10 ounces indicated on the container Critical value = 2.064 Since the test statistic 2.1742 is greater than the critical value 2.064, we reject the null hypothesis that true mean net weight is lower than the net weight of 10 ounces indicated on the container. The test is valid if there are no outliers and n is at least 15. Question 2 H0: H1: n = 100 P-value =- Critical value = 1.645 (i) The p-value method Since the p-value is greater than 0.05, we fail to reject the null hypothesis that the true proportion, p, of cola drinkers that prefer Coca-Cola over Pepsi is equal to 0.50. (ii) The rejection point method Since the z-statistic is less than the critical value, we fail to reject the null hypothesis that the true proportion, p, of cola drinkers that prefer Coca-Cola over Pepsi is equal to 0.50. Question 3 n1 = 15 n2 = 16 Ho: µ1-µ2 = 0 H1: µ1-µ2 ≠ 0 Sp = sqrt(((15-1)*3.4^2+(16-1)*3.8^2)/(31-2)) = 3.6124 Degrees of freedom = (31-2) = 29 p-value =- Critical value = 2.04523 (i) The p-value method Since the p-value is less than 0.05, we reject the null hypothesis that µ1-µ2 = 0 in favor of the alternative hypothesis. Therefore, there is a difference in the average score between the two methods. (ii) The rejection point method Since the test statistic is less than the critical value, we reject the null hypothesis that µ1-µ2 = 0 in favor of the alternative hypothesis. Therefore, there is a difference in the average score between the two methods. Question 4   Old Method New Method 36 29 55 42 28 30 40 32 62 56 Mean 44.2 37.8 Variance- n1 = 5 n2 = 5 Ho: µ1-µ2 = 0 H1: µ1-µ2 > 0 Sp = sqrt(((5-1)*195.2+(5-1)*130.2)/(10-2)) =- Degrees of freedom = (10-2) = 8 Critical value =- Since the test statistic is less than the critical value, we fail to reject the null hypothesis at 5 %. Therefore, the new method does not decrease the average length of the surgical procedure. Question 5 H0: H1: n = 100 Critical value = 1.645 Since the test statistic is greater than the critical value, we reject the null hypothesis in favor of the null hypothesis. Therefore, there is sufficient evidence to indicate that the new quality control standards were effective in reducing the proportion of defective digital music players. Question 6 a) In none 100 of these hypothesis tests did we reject Hₒ. In 50 of these hypothesis tests did I expect to reject Hₒ. b) All the p-values are all the same for the 100 tests because the 100 samples have the same mean and standard deviation. c) In general, the number of rejections increase if α = 0.01 is used instead of α = 0.10. d) Once again, using α = 0.10, in none of the tests did I reject the null hypothesis Hₒ: μ = 63.8. In 100 of these hypothesis tests did I expect to reject Hₒ. e) The rejection of Hₒ in part (A) is not a correct decision. Therefore, rejecting it would be a Type 1 error. f) The rejection of Hₒ in part (D) is not a correct decision. Therefore, rejecting it would be a Type 1 error.