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 = 20 , s = 4 , n = 32 , H O : μ = 22 , H a : μ &lt; 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  =  20  ,  s  =  4  ,  n  =  32  ,  H  O  :  μ  =  22  ,  H  a  :  μ  &amp;lt;  22

Amy Daniels · Accepted Answer

The test hypotheses are given below:  Null hypothesis:  H  0  :  μ  =  22  ,  Alternative hypothesis:  H  a  :  μ  &amp;lt;  22.Critical value approach:Critical value:For 5% level of significance,  α  =  1  −  0.95  α  =  0.05Hence, the cumulative area to the left is,Area to the left = 1 - area to  the right =1 - 0.05=0.95From Table of the standart normal distribution, the critical value is 1.64Test statistic:  z  =            x      −      μ              σ              n              =            20      −      22              4              32              =  −  2.83Decision  i  f  =&amp;lt;  −      =    a    ,  reject the null hypothesis       H    0  Conclusion:Here critical value is –1.645.Therefore, the test statistic is less than the critical value.For left-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, sample standard deviation, and sample size. In

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