Question

A researcher takes sample temperatures in Fahrenheit of 19 days from St. Louis and 17 days from Hartford. Test the claim that the mean temperature in St. Louis greater than the mean temperature in Hartford. Use a significance level of a = 0.10. Assume the populations are approximately normally distributed with unequal variances. You obtain the following two samples of data. Round answers to 4 decimal places. Hartford 61.9 89.3 44 St. Louis 85.9 94.7 85 84.3 90.6 85.7 80.2 74.8 92.5 76.3 83.9 80.2 83.7 96.9 85 87.2 78.8 76.7 85.6 81.3 91.2 78.6 108.3 38.6 63.1 99.9 64.7 60.7 84.3 52.3 64.2 70.1 66.2 What are the correct hypotheses? Note this may view better in full screen mode. Select the correct symbols for each of the 6 spaces. Ho: Hi Hj: CHI = 4MzA > 4 Hz -

Answer 1

in If the temperature in St. Louis and temperature Hartford be represented by the variables x and 4 respectively then : Null hypothesis , Ho: Me = My Hi: Melly (Right tailed) Under Ho, the test statistic is t= a- - ~ tnien- sekto) where so I [Elang & El-nada] nitn2-2 Now, n = 19, 2=17 5, = 84.63158 X= 71.68824 82 = [ (x-7)* + E140-1)] 19+17-2 [663.7611 +5791.6576] 34 - zu [ 6455.41877 - 189.86526 X2 ทเทจ 19t17 mee ting 19.717 - 382 = 0-1455 1) 455 nu n, 22 19 7

Hence t = 84.63158 - 71.68824 1189.86526 X 0.111455 12.94334 = 2.8 136717 4.60016 Tabulated to go for 34 d-f. for single tail test is 1.306952 and evalue is 0.004041 Now, since calculated t is greater than tabulated t value ,so, we conclude that the null hypothesss may not be accepted at a=0.10. since, the p value = 0.00404 <0.10. So, we conclude that the mean demperature in St. Louis is greater than the mean temperature in Hartford.