Question

S The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out to see whether this is the case. What conclusion is appropriate in each of the following situations? (a) n =20,t = 1.59, a = 0.05 Reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in O Reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in. Do not reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in. Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in. (b) n = 20, t= -1.59, a = 0.05 Reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in. Reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in. Do not reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in. O Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in. (c) n = 22, t = -2.53, a = 0.01 Reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in. O Reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in. Do not reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in. O Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in. (d) n = 22, t = -3.95 O Reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in. Reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in. O Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in. O Do not reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in. You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

ce Solutions de tere trece average diawala of ball bearings of castain lippe According to the question The 0.5 nus hull hypothesis, Hoi pe= The allemnelige lypellosis, H4040.5

la Given n = 20 n t = 1.59 and X = 0.05 thena Csilal value o -1) Hoogs, 19) - 2093 dea = £19) 2.5% -2.093 2.093 Heren lel< (len so accept to Auswer Donot rejech the null lypo Dhesis These is not suppricut evidence that the toue diamelos offers from 0.5 lu

(6 Given, n=20 t = -1.59 and X=0.05 Thena Csilical value o .n-1) " 40028, 19) = 2.093 419) 2.5% 2.5% -2.093 2.093 Heren lel< lfcln so accept to. Donot xojech the null lypo Thesis There is not saffrieuf evidence their the true diamela offers from 0.5 lu Auswer

Given, n=2a, t= -2.53 and a=0.01 Theur Critical valle. de = hann (0.005,21) - 2.831 421) 9:59% 0.5% 2.831 -2.031 ikeres [l <1841 ci so Accept Ho Answer: Do not rejecit tre null hypothesis, Hier is not sufficient eviance the true diametek Cuffers Jom 0.5 lin.

2) Gurkeur n=, f:-3.95 and x=0.01 Then critical value dc= teorie h-)) +0.00521) = 9.831 E 0.5% 1.5% 2031 2.831 Here lllz llen. So wjeet Ho. Answer: Rejech The null hypothesis These is suffisant evidence that toue oicuneta differs fom 0.5 in.