Recall: Ch.23,25,and 24 we did CI (confidence intervals) Ch. 23 was the single mean, Ch. 25 was paired samples, and Ch. 24 was independent samples. All of these used the student’s t-model. Hypothesis Test Ex. For a single mean with a hypothesis test. # 37 on pg. 614 =Weight of vehicles. The auto manufacturers and insurance companies believe the avg. weight of a U.S. made car is 3000 lbs. Random sample of 91 U.S. cars, the avg. weight being 2919 lbs with a SD = 531.5 lbs. Is there strong enough evidence to claim the mean weight of all U.S. cars is different than 3000 lbs. tn-1 = y(hat) – mean/ SE(y(hat)) H⁰ = mean (mu) = 3000 lbs SE(y(hat)) = SD/square root of n Ha = mean (mu) not equal to 3000 lbs. y(hat) = 2919 your sample mean SD = 531.5 lbs n = 91 SE(y(hat)) 531.5/square root of 91 = 55.71 DOF = 91-1 = 90 degrees of freedom tn-1 t90= 2919 – 3000/ 55.71 = -1.454 p-value = 0.1495 because alpha level is 0.025 so this p-value is high. So we fail to reject the null hypothesis. Therefore, we retain the null hypothesis and conclude the mean weight of U.S. cars to be 3000 lbs. Ex. Paired samples – null hypothesis for paired samples is always the same. The mean of the differences Md = 0. Ways to reduce gas millage. Work 4 – 10 hour days instead of 5 – 8 hour days. A company is worried about gas prices so they switch to the above work schedule to see if they drive less miles. Is there evidence a 4- day work week would change amount of miles driven. 5-day work week = 2798 miles, 7724 miles… tn-1 = d(bar) – 0/ SE(d(bar)) 4- day work week =2914 miles,6112 miles… SE(d(bar)) = Sd/square root of n 1139.6/square root of 11 = 343.6 Take 5-day work week subtract from 4- day work week. DOF = 11-1 = 10 So we have the differences as follows -166, 1612… t10 = 982/ 343.6 = 2.86 = # of SD away from mean. Alpha level is 0.025 p-value = 0.017 So we can reject the null hypothesis and conclude that a change in work week did lead to a different average millage. Summary n = 11 employees D(bar) = 982 miles (avg difference in miles driven) Sd = 1139.6 miles Ho: Md = 0 and Ha: Md not equal to 0 If this is a one tail test alpha is 0.05 if it’s a two tail test it alpha level will be 0.05/2 = 0.025 If alternative Ha is greater than or less than it’s a one tail test if it’s not equal to it’s a two tail test. Ex. Independent samples: Ho = M1 – M2 = 0 always this unless told differently Listening to music while you study. A class of stat students randomly assigned 62 people to either listen to Mozart or to rap music while studying for a certain test. Does it appear Mozart is better than rap while studying for a test? Summary – Mozart = 20 students avg. score on test was 10.00 and SD was 3.19 Rap = 29 students avg. score was 10.72 and SD was 3.19 Ho = mean of Mozart – mean of Rap music = 0 Ha = Mean of Mozart – mean of rap music > 0 t = (y(bar)1 – y(bar)2) – 0/ SE(y(bar)1 – y(bar)2) SE(y(bar)1 – y(bar)2) = square root of S12/n1 + S22/n2 Square root of (3.192/20) + (3.192/29) = 1.02 T = (10-10.72)-0/1.02 = -.71 df= 45.88 alpha = 0.05 p-value = 0.7563 Due to large p-value of .7563 we fail to reject null hypothesis and conclude that there is no evidence that Mozart is better than rap to listen to while studying.