Asample of 47 observations is selected from a normal population. the sample mean is 30, and the population standard deviation is 5. conduct the following test of hypothesis using the 0.05 significance level. h0 : μ ≤ 29 h1 : μ > 29

a. is this a one- or two-tailed test? "two-tailed"-the alternate hypothesis is different from direction. "one-tailed"-the alternate hypothesis is greater than direction.

b. what is the decision rule? (round your answer to 3 decimal places.) h0, when z > 1.645

c. what is the value of the test statistic? (round your answer to 2 decimal places.) value of the test statistic 1.37

d. what is your decision regarding h0? do not reject reject there is evidence to conclude that the population mean is greater than 29.

e. what is the p-value? (round your answer to 4 decimal places.) p-value.

We accept the null hypothesis and reject  the alternate hypothesis. There is no evidence to conclude that the population mean is greater than 29. The population mean is less than or equal to 29.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 29

Sample mean, = 30

Sample size, n = 47

Alpha, α = 0.05

Population standard deviation, σ = 5

First, we design the null and the alternate hypothesis

a) This is a one-tailed test because the alternate hypothesis is in greater than direction.

We use One-tailed z test to perform this hypothesis.

b) , we reject the null hypothesis and accept the alternate hypothesis and if , we accept the null hypothesis and reject  the alternate hypothesis.

c) Formula:

Putting all the values, we have

d) Now,

Since,  

We accept the null hypothesis and reject  the alternate hypothesis. There is no evidence to conclude that the population mean is greater than 29. The population mean is less than or equal to 29.

e) P-value is 0.0853

On the basis of p value we again accept the null hypothesis.

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