Question

5. You are given the following information about a hypothesis test that is being performed on a random sample taken from a normal distribution with mean : • The unbiased sample variance is 9. • The sample size is 16. • H:= 80 • He: H > 80 Let tay be the critical value of a t random variable with v degrees of freedom. The following table lists values of ta for specific combinations of a and v: v = 15 v = 16 a = 0.01a = 0.005 2.6022.947 2.583 2.921 Calculate the highest possible sample mean that would result in the null hypothesis not being rejected at the 1% significance level.

Answer 1

The unbiased sample variance is sº = 9 The sample size is n = 16 The level of significance is a =0.01. The population mean is klo = 80). The corresponding t - distribution and the critical value with a=0.01 is obtained by using the Excel 2010 function is = tinv(probability degrees of freedom) tiv(0.01,15) = 2.60248 600L1s = 12.602 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - The determination of the sample mean(x): The highest possible sample mean that would result in the null hypothesis not being rejected at the 1% significance level is =X=hous (fr) + =X=260248(26) +20 =X =2.60248(4) +80 = X = 2.60248(0.75) +80 1.95186+80 = 81.95186 X 81.95 Therefore, the highest possible sample mean that would result in the null hypothesis not being rejected at the 1% significance level is X 81.95