Question

A sample mean, sample standard deviation, and sample size are given. Use the one meant test to perform the required hypothesis test about the mean, , of the population from which the sample was drawn. Use the P-value approach. Also, assess the strength of the evidence against the null hypothesis. x-22,298, s=14200, n = 17, HO: P = 30,000, Ha# 30,000 a -0.05. Test statistic: 224. P.value 0.0200. Reject the null hypothesis. There is sufficient evidence to conclude that the mean is different from 30,000 The evidence against the null hypothesis is strong Test statistic: t = -2.24. P-value = 0.0399. Reject the null hypothesis. There is sufficient evidence to conclude that the mean is different from 30,000 The evidence against the null hypothesis is strong Tost statistic: 224. P value 0.9601. Do not reject the null hypothesis. There is not sufficient evidence to conclude that the mean is different from 30,000. The evidence against the null hypothesis is weak or none Test statistic -2.24. P-value - 0.0250. Reject the null hypothesis. There is sufficient evidence to conclude that the mean is different from 30,000 The evidence against the null hypothesis is strong

Answer 1

# givers that," F = 22,298 58 14 20 0 n=17 Hypothesis is Hoi 1=30000 us Hoi M £30000 daoios t test: Use One-Sample meau talubatan - taluated X-M s/n = 22,290- 30000 14200 = -7702 Calculated t-value - 3uuuool =-2.236 ₂-2.24 now Now or is 5% a degree of freedom (dif) = n-1 = 1741 = 16 from t-table of two tuiled ? Tabulated to =2.120 5,16) ( from pred p-value calculcites] P-value = 0.0399 So, and option is correct. to-2:24 P-value = 0.0399 , Reject null Mypothesis. there is sufficient evidicule to conclude that the mean is different from 30000. please give up reate if it is helpful to you,