Question

The frequency distribution shows the results of 200 test scores. Are the test scores normally distributed? Use a =0.01. Complete parts (a) through (e). Class boundaries 49.5-58.5 58.5-67.5 Frequency, f 19 62 D 67.5-76.5 81 76.5-85.5 33 85.5-94.5 5 Using a chi-square goodness-of-fit test, you can decide, with some degree of certainty, whether a variable is normally distributed. In all chi-square tests for normality, the null and alternative hypotheses are as follows. Ho: The test scores have a normal distribution. Ha: The test scores do not have a normal distribution. To determine the expected frequencies when performing a chi-square test for normality, first find the mean and standard deviation of the frequency distribution. Then, use the mean and standard deviation to compute the z-score for each class boundary. Then, use the Z-scores to calculate the area under the standard normal curve for each class. Multiplying the resulting class areas by the sample size yields the expected frequency for each class. (a) Find the expected frequencies. 85.5-94.5 Class boundaries Frequency, f Expected frequency 49.5-58.5 19 58.5-67.5 62 67.5-76.5 81 76.5-85.5 33 (Round to the nearest integer as needed.) (b) Determine the critical value, Xo, and the rejection region. x. =(Round to three decimal places as needed.) Choose the correct rejection region below. O A x?sx OC.x2x2 OB. *?<x? OD. x>x? (c) Calculate the test statistic. x=(Round to three decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. V Ho

Answer 1

(a) E; = np: = 200 བ(3) 5 = 40 The expected value for each value is 40. 58.5-67.5 67.5-76.5 76.5-85.5 85.5-94.5 Class 49.5-58.5 boundaries Frequency,f 19 Expected 40 frequency 62 40 81 40 33 40 5 40 The critical value is x3 =13.277. Rejection region:

D. x>x (c) (0;-E) χ =Σ E (19-40)2 (62 - 40) (81-40) (33 - 40) (5-40)? 40 40 40 40 40 = 11.025 +12.1+ 42.025 +1.225 +30.625 = 97.000 + + + + The value of the test statistic is x2 =97.000. Reject H -