Testing Goodness-of-Fit with a Normal Distribution? Refer to Data Set 1 in Appendix B for the 40 heights of females. G r L 1 1 e e 5 6 a s 5 2 H t s . . e e T 4 0 i r h 1 0 g T a 0 5 h h n - - t a 1 1 1 ( n 5 6 6 c 1 5 2 8 m . . . 6 ) 8 4 0 6 . 1 0 0 6 0 5 1 0 1 F r e q u e n c y a. Enter the observed frequencies in the preceding table. b. Assuming a normal distribution with mean and standard deviation given by the sample mean and standard deviation, use the methods of Chapter 6 to find the probability of a randomly selected height belonging to each class. c. Using the probabilities found in part (b), find the expected frequency for each category. d. Use a 0.01 significance level to test the claim that the heights were randomly selected from a normally distributed population. Does the goodness-of-fit test suggest that the data are from a normally distributed population?

Solution 25BSC Step 1 a. The observed frequencies is given in the preceding table. 155.410-162. 162.006-168.6 Greater than Height (cm) Less than 155.410 006 01 168.601 Frequency 6 13 15 6 b. By assuming a normal distribution with mean and standard deviation given by the sample mean and standard deviation, by using the methods of Chapter 6 to find the probability of a randomly selected height belonging to each class. Now, the probability of randomly selected height belonging to each class is 0.159, 0.341, 0.341, 0.158. c. The expected frequency for each category is shown below 155.410-162. 162.006-168.6 Greater than Height (cm) Less than 155.410 006 01 168.601 Frequency 6 13 15 6 Probability 0.159 0.341 0.341 .158 Expected frequency=probabi 6.348 13.652 13.652 6.348 lity*40 d. By using = 0.01 significance level to test the claim that the heights were randomly selected from a normally distributed population. We are testing Chi-square test (using EXCEL) as shown below Goodness of Fit Test expecte 2 observed (O - E) (Oi Ei) d Ei 6 6.348 -0.348 0.019 13 13.652 -0.652 0.031 15 13.652 1.348 0.133 6 6.348 -0.348 0.019 chi-squa 0.20 re df 3 P-value 0.9772 Now, we have = 0.20 is the obtained test statistic value. Here, k is the number of different categories of outcomes = 4 Then degrees of freedom = k - 1 = 4 - 1 = 3 and P-value = 0.9772 (The smaller P-value is, the stronger the evidence against H and in favo0of H . If P-value is1 small like 0.01 or smaller, we may conclude that the null hypothesis H is strongly rejected in 0 favor of H .1f P-value is between 0.05 P-value 0.01, we may conclude that the null hypothesis H is rejected in favor of H . In other cases, i.e., P-value > 0.05, we may conclude that 0 1 the null hypothesis H is 0cepted) Since P-value is greater we accept the null hypothesis at 1% level of significance and conclude that there is evidence to claim that the heights were randomly selected from a normally distributed population. Hence the goodness-of-fit test suggest that the data are from a normally distributed population.