Question

In a current study, a random sample of 20 teachers from Texas and an independent random sample of 20 te have been asked to report their annual income. The data obtained are as follows: Annual income in dollars 35732, 37959, 43098, 21621, 40725, 27525, 27542, 43119, 42290, 48018, 36302, 41585, 37961, 44740, 35775, 27635, 45494, 33006, 50627, 45047 41760, 41471, 43483, 48834, 36943, 30975, 44291, 41084,35323, 42696, 45377, 29561, Texas indan42903, 46986, 39003, 30290, 41588, 36271, 38546, 51810 to Excel The population standard deviations for the annual incomes of teachers in Texas and in Indiana are estimated as $620 respectively. It is also known that both populations are approximately normally distributed. Construct a 90% confiden for the difference μι-μ2 between the mean annual income of teachers from Texas (ul) and the mean annual income from Indiana (u2). Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal pla necessary, consult a list of formulas.) What is the lower limit of the 90% confidence interval? What is the upper limit of the 90% confidence interval?

Answer 1

Solution:-

Lower limit of the 90% confidence interval is - 5761.945 .

Upper limit of the 90% confidence interval is 1422.545 .

90% confidence interval for the difference between mean annual income of teachers from Texas and Indiana is C.I = ( - 5761.945, 1422.545)

Texas Indiana Sum Mean S.D Count S.E 765801 809195 38290.05 40459.75 7722.0244787 5978.660322 20 1726.6971658 1336.869089 20

$C.I = \left (\bar{x}_{1}-\bar{x}_{2} \right )\pm z_{\alpha /2}\times \sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}$

C.I = (38290.05 - 40459.75) + 1.645*2183.635805

C.I = - 2169.7 + 3592.2454

C.I = ( - 5761.945, 1422.545)