Question

plz this question was answer for me yesterday but i an difficulties in figuring out the right answer. i will be glad if u can put my final answer on the bix ao i can easy figure it out. thanks

12. Preliminary data analyses indicate that you can reasonably consider the assumptions for using pooled t-procedures satisfied. Independent random samples of released prisoners in the fraud and firearms offense categories yielded the following information on time served in months. Obtain a 95% confidence interval for the difference between the mean times served by prisoners in the fraud and firearms offense categories. (Note: X = 13.63, S, = 3.95, X, = 19.94, and S, = 4.82.) Fraud 16.3 16.8 11.6 9.3 19.3 15.9 11.1 7.7 10.9 17.4 Firearms 18.8 23.6 25.8 25.9 16.9 15.8 25.2 13.1 15.1 19.2 The 95% confidence interval is from (Round to three decimal places as needed.) 13. Provided below are summary statistics for independent simple random samples from two populations. Use the nonpooled t-test and the nonpooled l-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval. Xy - 10,54 = 4, n = 10.x2 = 13, S2 = 8, 12 = 10 a. Two-tailed test,t=0.01 b. 99% confidence interval a. What are the hypotheses for the t-test? O A. Ho - - P2 Ho Rih Oc. Họ Ha = 92 H₂ Hith OB. Ho 1:22 ни на OD. HS # 2 HH2 Find the test statistic (Round to three decimal places as needed.) Find the critical values. 12 + (Round to three decimal places as needed.) What is the condusion of the hypothesis test? O A Reject He There is sufficient evidence that the two means are different. O B. Do not reject Ho. There is sufficient evidence that the two means are different O C. Rent There is insufficient evidence that the two means are different. OD. Do not reject Hy. There is insufficient evidence that the two means are different. b. The 90% confidence interval is from (Round to three decimal places as needed.)

Answer 1

Answers 12] , = 13.63, 712 = 19,94 5 ,= 3.95, S2=4.82 = 21.5749 Sun, S ms - 1013.95 + 10 (482) nina-2 lot1o-2 S= 4.65 tnt-2.0.025 = 2.445 - 95%. Confidence interval is ( 51 - Th - tn.11222,0,025 s htator T2+th, in 2,9025 x st) => (-631-5.085, -6.31+5.085) 3) (-11.395, -1.225) to -1,225 The 95% confidence interval is from ) - 11.395 13] - » c. Ho : M, = M2 Haibli & Mr Test statistic, + = -2 S = 6.067 s=1n.s? + N252 V7+92-2 ; t = 10-13 0.007 2 to-1.006 Critical value i tava = † 3.1977

conclusion in D] Do not reject Ho. There is |_means are different insufficient evidence that the two 67 99% confidence interval is, (7,-ve- taiz sjhitte, 71, -2a + tarz Sy het med (-3-9.532,-3+5.532) (-12.532, 6-532) - The 90% confidence interval is from 12.532] to 16.532)