The frequency distribution shows the results of 200 test scores. Are the test scores normally distributed? Use a= 0.05. Complete parts (a) through (e). Class boundaries Frequency, 49.5-58.5 58.5-67.5 67.5-76.5 76.5-85.5 85.5-94.5 20 60 82 34 4 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. H: 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. 49.5-58.5 20 58.5-67.5 67.5-76.5 76.5-85.5 34 85.5-94.5 Class boundaries Frequency, f Expected frequency 60 82 (Round to the nearest integer as needed.) (b) Determine the critical value, Xo- and the rejection region. xb =[ Round to three decimal places as needed) Choose the correct rejection region below. OB. x Oc sx (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. Ho. (e) Interpret the decision in the context of the original claim. At the 5% significance level, there V enough evidence to reject the claim that the test scores are normally distributed.
The frequency distribution shows the results of 200 test scores. Are the test scores normally distributed? Use a= 0.05. Complete parts (a) through (e). Class boundaries Frequency, 49.5-58.5 58.5-67.5 67.5-76.5 76.5-85.5 85.5-94.5 20 60 82 34 4 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. H: 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. 49.5-58.5 20 58.5-67.5 67.5-76.5 76.5-85.5 34 85.5-94.5 Class boundaries Frequency, f Expected frequency 60 82 (Round to the nearest integer as needed.) (b) Determine the critical value, Xo- and the rejection region. xb =[ Round to three decimal places as needed) Choose the correct rejection region below. OB. x Oc sx (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. Ho. (e) Interpret the decision in the context of the original claim. At the 5% significance level, there V enough evidence to reject the claim that the test scores are normally distributed.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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