Question

Past records suggest that the mean annual income, 12, of teachers in state of Georgia is greater than or equal to the mean annual income, H2, of teachers in Indiana. In a current study, a random sample of 15 teachers from Georgia and an independent random sample of 15 teachers from Indiana have been asked to report their mean annual income. The data obtained are as follows. Annual income in dollars Georgia 35563, 41690, 51807, 41824, 35509, 35642, 31758, 46318, 34484, 41247, 35647, 30176, 46458, 37161, 38522 Indiana 39610, 38810, 36678, 41502, 54314, 35645, 43834, 49092, 40356, 33131, 35058, 41962, 44810, 44635, 43976 Send data to Excel The population standard deviation for mean annual income of teachers in Georgia and in Indiana are estimated as 6500 and 6600, respectively. It is also known that both populations are approximately normally distributed. At the 0.05 level of significance, is there sufficient evidence to reject the claim that the mean annual income of teachers in state of Georgia is greater than or equal to the mean annual income of teachers in Indiana? Perform a one-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. р a р The null hypothesis: x s ke The alternative hypothesis: O=O OSO DO The type of test statistic: (Choose one) <D D> The value of the test statistic: (Round to at least three decimal places.) 0 х 5 ? The critical value at the 0.05 level of significance: (Round to at least three decimal places.) Can we reject the claim that the mean annual income of teachers from Georgia is greater than or equal to the mean annual income of teachers from Indiana? Yes O No

Answer 1

Bookl - Microsoft Excel File Home Insert Page Layout Formulas Data Review View Cut Calibri 11 Α Α΄ Wrap Text General Copy Paste B 1 U T А Merge & Center Y % 4.0 .00 .00 0 Format Painter Clipboard Font Alignment Number R21 fic A B D E F G H J K 1 N 3 4 5 6 7 8 9 с Indlana x2 39610 38810 36678 41502 54314 35645 43834 49092 40356 33131 35058 41962 44810 44635 43976 623413 41561 Georgia x1 35563 41690 51807 41824 35509 35642 31758 46318 34484 41247 35647 30176 46458 37161 38522 583806 38920.4 10 11 12 13 14 15 16 17 18 Total 19 Average

Answerin > Here the null hypothesis Ho: Moyua Alternative hypothesis Hi: M, LU2 Here the porudation standard deviations for the two-samples has givere, so, we need to use the z- distribution Griven data: M = 15 X, = 38920.4 ar 6500 na : 15 Xa=41560.96 na=660o + Tegl statistic x - Xa 7 1 12 (38920.k 41566.86) (6500)? (6600)? + 15 15 -1.103 We use here we are using 7. distributiore! when use 7-test ro rieed to calculate degrees of of fredom we be calculated when to distributions, freedom. Degree we use r 2=0.05, critical value 1.645 → calz= -1.103 > -1.645 So, donot regect Ho con't reject the claim NO,