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Hypothesis test for the mean: For our purposes, we will be doing a T-test on the mean and an unpooled test when testing the difference between two means. Test for one sample mean Judy is a ad designer who designs the newspaper ads for the Giant grocery store. Electronic counters at the entrance total the number of people entering the store. Before Judy was hired, the mean number of people entering every day was 3018. Since she has started working at the Giant the management thinks that this average has increased. A random sample of 42 business days gave an average of 3333 people entering the store daily with a standard deviation of 287. Does this indicate that the average number of people entering the store every day has increased? Use an alpha of 0.01. Stat…. Tests…T-Test Inpt: choose data or stats according to what is available (usually you use stats) 0 : stands for the hypothecated mean which is in your hypothesis In this case we are testing against the previous 3018. X bar is the sample mean of 3333 Sx is the standard deviation of 287 n: the number in your sample 42 days : choose the notation used in the alternative hypothesis In this case we are looking for an increase so choose > Calculate We get a p value of .000000006 which is very, very slight. Therefore, we reject the null hypothesis and conclude that the average number of people entering the store has certainly increased. Test for two samples According to a survey by the Road Information Program, Ohio is the pothole capital of the United States with an estimated 6.8 million potholes. Potholes are caused by water seeping into cracks, freezing and swelling the pavement. A measurement is devised to test the resistance to water. From the following data, determine whether there is significant difference in the resistance to water for the two types of materials. Alpha = .05 Material A Material B n 36 40 mean 63.8 87.2 Stat…Tests…2-SampTTest…Stats standard deviation 16.66 35.75 X1 is 63.8 and sx1 is 16.66 and n1 is 36 X2 is 87.2 and sx2 is 35.75 and n2 is 40 Next you choose the symbol used in the alternative hypothesis. Since we are testing for a significance difference, that is asking if they are the same or not. So, choose the not equal option. DO NOT pool Calculate You should get a p value of .00047 which means you would reject the null hypothesis which states the two means are equal and conclude that there is a difference in the materials. Test for matched pair (dependent sample) Figure Perfect, Inc. is a women's figure salon that specializes in weight-reduction programs. Weights in pounds for a sample of clients before and after a 6-week introductory program are as follows. Client 1 2 3 4 5 6 Weight before 140 160 210 148 190 170 Weight after 132 158 195 152 180 164 Use an alpha of 0.05 to determine if people lost weight. If the program works, the weights before the program began would be higher than the weights after the program. So the null hypothesis is that the “before” weights = the “after” weights and the alternative hypothesis is that the “before” weights > “after” weights meaning the program was successful. Another way to look at the alternative is that before – after >0. Put the before weights in L1 and the after weights in L2. Go to the top of L3 and type L1 – L2 then enter. Now do a T test on L3. mu sub 0 is 0; use L3 for the list; the freq is 1; choose > since if the weights before the program are significantly greater tean the after weights, the program worked. Calculate. The p value is .0352 which is lower that .05 so the program did help people lose weight. So, we reject the hypothesis and conclude that there is a difference. However, at the .01 significance level, the program did not work.