# A sample of 36 observations is

A sample of 36 observations is selected from a normalpopulation. The sample mean is 21, and the population standarddeviation is 5. Conduct the following test of hypothesis using the0.05 significance level.

H0: μ ≤ 20

H1: μ > 20

a). What is the decision rule? (Round your answer to 2decimal places.)

b). What is the value of the test statistic? (Round youranswer to 2 decimal places.)

Solution :

= 20

= 21

= 5

n = 36

This is the righttailed test .

The null andalternative hypothesis is ,

H0:   ≤ 20

Ha : > 20

The significance levelis α=0.05,

The critical value fora right-tailed test is zc​=1.64.

The rejection regionfor this right-tailed test is R = (z:z >1.64)

Test statistic = z

= ( ) / / n

= (21 – 20) /5 /36

= 1.20

P(z >1.20 ) = 1 -P(z < 1.20 ) = 0.1151

P-value = 0.1151

= 0.05

p=0.1151≥0.05, it isconcluded that the null hypothesis is not rejected.

There is not enoughevidence to claim that the population meanμ is greater than 20, atthe 0.05 significance leveL

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