Question

Use the traditional method to test the given hypothesis. Assume that the population is normally distributed and that the sample has been randomly selected. 36) A machine dispenses a liquid drug into bottles in such a way that the standard deviation of 36) the contents is 81 milliliters. A new machine is tested on a sample of 24 containers and the standard deviation for this sample group is found to be 26 milliliters. At the 0.05 level of significance, test the claim that the amounts dispensed by the new machine have a smaller standard deviation 37) 37) At the a= 0.05 significance level test the claim that a population has a standard deviation of 20.3. A random sample of 18 people yields a standard deviation of 27.1.

THERE IS QUESTION AND ANSWERS FOR #36 AND #37.... I HAVE GIVEN U THE ANSWERS....I JUST NEED THE STEP BY STEP SOLUTION.

Answer 1

Step1: The summary statistics: The sample size is n = 24, The population standard deviation is o = 81 milliliters. The sample standard deviation is s= 26 milliliters, State the null and alternative hypotheses. H,: 0=81 (The population variance is equal to 81) versus H,: 0 <81 (The population variance is less than 81) Step 2: The level of significance is a=0.05. Select the distribution to use. We use the chi-square distribution to test a hypothesis about o". --------------------------- -- ---- ------ -- Step3: Determine the rejection and non-rejection regions. The significance level is 5%. The <sign in H, indicates that the test is left-tailed. The degrees of freedom are df = n-1 = 24-1 = 23 df = 23 The critical value of 72 for 23 degrees of freedom and for a=0.05 area of the right-tail is z*for 23 df and 0.05 area in the right tail = [13.09051] - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Step4: Calculate the value of the test statistic. The value of the test statistic za for the sample variance is calculated as follows: x2 = (n=1) (24-1) 676 6561 23(676) 6561 15548 6561 = 2.369761 72 = 2.3697 -------- - - - - - - The probability value of the chi-square test statistic is The P-value approach: P(x2 > 2.3697)=1- P(x* < 2.3697) = 1-0.000 >1 P(x* > 2.3697) 1 > a = 0.05). We do not reject the null hypothesis H. ----- ------ -------- -------- ---------- ------ -- Step 5: Make a decision Since, the probability value is P-value = 1 > a = 0.05 We do not reject the null hypothesis Ho. The value of the test statistic is za = 2.3697 < xã 0323=38.07563 Consequently, we do not reject the null hypothesis H. Hence, we conclude that there is no evidence that the population variance is less than 81 milliliters at the 5% level of significance.