In each of Exercises, we have provided a sample mean, sample standard deviation, and sample size. In each case, use the one-mean t-test to perform the required hypothesis test at the 5% significance level.x = 24, s = 4, n = 15, H0: μ = 22, Ha: μ &gt; 22

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In each of Exercises, we have provided a sample mean, sample standard deviation, and sample size. In each case, use the one-mean t-test to perform the required hypothesis test at the 5% significance level.x = 24, s = 4, n = 15, H0: μ = 22, Ha: μ &amp;gt; 22

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The test hypotheses are given below:Null hypothesis:H0:μ=22,Alternative hypothesis: Ha:μ&amp;gt;22.Critical value approach:Critical value:For 5% level of significance,α =1-0.95α = 0.05Hence, the cumulative area to the left is,Area to theleft = 1- Area to theright=1-0.05= 0.95From Table of the standard normal distribution, the critical value is 1.645.Test statistic:z=x−μσn=24−22415=1.94If test statistics is greater than the critical value, then reject the null hypothesis. Otherwise, do not reject the null hypothesis.Conclusion:Here, the test statistic is greater than the critical value.For right-tailed test, the null hypothesis is rejected.Thus, it can be said that there is sufficient evidence to reject the null hypothesis.

In each of Exercises, we have provided a sample mean,

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