Question

View Policies Show Attempt History Current Attempt in Progress Use the t-distribution to find a confidence interval for a difference in means – Hly given the relevant sample results. Give the best estimate for 44, - Hy, the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed. A 90% confidence interval for #, – H, using the sample results xı = 502, 51 = 120, n= 360 and x2 = 435, 52 = 94, n2 = 200 Enter the exact answer for the best estimate and round your answers for the margin of error and the confidence interval to two decimal places. Best estimate = i 671 Margin of error = Confidence interval: i to i Save for Later Attempts: 2 of 4 used Submit Answer

Answer 1

Point Estimate

$\( \bar{X}_{1} - \bar{X}_{2}) = 502 - 435 = 67$

Margin of Error = $t_{\alpha /2 , DF}\sqrt{ (S_{1}^{2}/n1) + (S_{2}^{2}/n2)} = 15.12$

Confidence interval :-
$\( \bar{X}_{1} - \bar{X}_{2}) \pm t_{\alpha /2 , DF}\sqrt{ (S_{1}^{2}/n1) + (S_{2}^{2}/n2)}$
t(α/2, DF) = t(0.01 /2, 496 ) = 1.648

DF = 496

90% Confidence interval is ( 51.88 , 82.12 )