Question

The frequency distribution shows the results of 200 test scores Are the test scores nomally distributed? Use α-0.10 Complete parts is) through (e) Class boundaries Frequency, 49.5-585 2D 58.5-67.5 51 67.5-76.5 B0 76.5-85.5 35 95.5-94.5 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 nomality the null and alternative hypotheses are as follows Ho The test scores have a nomal distribution Ha The test scores do not have a normal distribution To determine the cxpected trequencics when performing a chi-square test for normality, first tind the mcan and standard deviaticn of the frequency distribution. Then, use the mean and standard deviation to compute the 2-sccre for cach class boundary Then use the 2-scores to calculate the area under the standard nomal Curve tor each dlass Multiplying the resulting cass areas by the sample size yields the expected trequency tor each class (a) Find the expected frequencies 58 5-67 5 61 lass boundaries 49 5-58 5 20 67 5-765 6 5-85 5 35 85 5-945 cted frequency (Round to the ncarest integer as needod.) (b) Determine te critical value, χ0, and the 'ejection legion. (Round to three decimal places as needed ) : Choose the correct rejection region below (c) Calculate the test statistic Click to select your answers). (b) Determine the critical value, χ, and the rejection region I l (Round to three decimal places as needed ) Choose the correct rejection region below (c) Calculate the test statistic. Round to three decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis e) Interpret the decisiorn in the contexd of the original claim At the 10% significance level, there ' enough evidence to elect the claim that the test scores are normally distributed.

Answer 1

Ans: clase 6 4 o 9.5-58. 20 11, 025 46 0,62S 32.4 4o 6 1 67.5-76.5 76.52.535 35.5-q4.5 니。 フ%, 94.05 Retct theless eto hethen thal the test Sanch