Stat 2606 E Assignment 4 (Statistics Project Sample)

Student’s Name Professor’s Name Course Date Stat 2606 E: Assignment 4 Question 1 n = 25 x=9.48 µ =10 σ=1.43 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 t = 9.48-101.4325 = 2.1742 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: p=0.5 H1: p>0.5 n = 100 p = 58100 =0.58 z = 0.58-0.50.5*0.5100 = 0.080.05 =1.6 P-value = 0.9452007 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 t = 69-72-03.6124*115+116 = -31.298299 = -2.3107 Degrees of freedom = (31-2) = 29 p-value = 0.01407901 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 195.2 130.2 n1 = 5 n2 = 5 Ho: µ1-µ2 = 0 H1: µ1-µ2 > 0 Sp = sqrt(((5-1)*195.2+(5-1)*130.2)/(10-2)) = 12.75539 t = 195.2-130.2-012.75539*25 = 3.222921 Degrees of freedom = (10-2) = 8 Critical value = 3.355387 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: p=0.25 H1: p<0.25 n = ...