please with excel formulas If n2 100, you may use normal distribution in all cases.1) Guidelines for the Jolly Blue Giant Health Insurance Company say that the averagehospitalization for a triple hernia operation should not exceed 36 hours. A diligent auditorstudied records of 16 randomly chosen triple hernia operations at Hackmore Hospital andfound a mean hospital stay of 40 hours with a standard deviation of 20 hours. "Aha!" shecried, "the average stay exceeds the guideline.". Using the .025 level of significance,conduct a Hypothesis test.a) State the Null and Alternate Hypothesis.b) Which test statistic does the data call for (which distribution)? Explain why you chosethat distribution.c) State the decision rule and Critical Value.d) Compute the Test Statistic and p value.e) What is your decision concerning null hypothesis?f) Re-compute the test statistic and p-value if “n" is increased to 121 using the samemean and standard deviation.g) What is your decision then?h) What is the minimum sample size if the allowable error, E, is 2 at .025 level ofsignificance.

As per our guidelines, we are allowed to answer first three sub-parts only. Thanks Hypothesized mean = 36 hours Sample size , n = 16 sample mean , x = 40 hours sample standard deviation, s = 20 hours a) Null and Alternative Hypotheses : H0:μ≥36 (The average hospitalization for a triple hernia operation exceed the guidlines)Ha:μ<36 (The average hospitalization for a triple hernia operation does not exceed the guidlines) b) It is a left-tailed test, for. We have sample size less than 30, also population standard deviation is not provided . We will use a t-test for one mean, with unknown population standard deviation Decision Rule : Significance level = α=0.025\alpha = 0.025 Degrees of freedom = n-1 = 15 Critical value : tc=−2.131t_c = -2.131 Rejection region : R={t:t<−2.131}R = \{t: t < -2.131\} Reject H0 : R={t:t<−2.131}R = \{t: t < -2.131\} Test Statistics : t=Xˉ−μ0s/n=40−3620/16=0.8 \begin{array}{ccl} t & = & \displaystyle \frac{\bar X - \mu_0}{s/\sqrt{n}} \\\\ \\\\ & = & \displaystyle \frac{ 40 - 36}{ 20/\sqrt{ 16}} \\\\ \\\\ & = & 0.8 \end{array} Decision : We noticed that t=0.8≥tc=−2.131t = 0.8 \ge t_c = -2.131 Hence null hypothesis is not rejected P-value using Excel : =T.DIST(0.8,15,TRUE) The p-value is p=0.7819p = 0.7819