# Question

Scores for a sample of 350 job applicants taking an aptitude test are summarized in the table below:

The average test score for the sample was 65. The standard deviation was 15. Conduct an appropriate chi square goodness of fit test to test a null hypothesis that test score follows a normal distribution. Use the mean sample score as your estimate of µ, the mean of the normal distribution that you will try to fit to the sample data. Use the sample standard deviation as your estimate of σ, the standard deviation of that normal distribution. Use a 5% level of significance. Compute degrees of freedom as c–3, where c is the number of test score categories. (The reason df is not computed as c 1 as shown in the chapter is that one additional degree of freedom is lost when the sample mean is used to estimate and one more degree of freedom is lost when the sample standard deviation is used to estimate)

The average test score for the sample was 65. The standard deviation was 15. Conduct an appropriate chi square goodness of fit test to test a null hypothesis that test score follows a normal distribution. Use the mean sample score as your estimate of µ, the mean of the normal distribution that you will try to fit to the sample data. Use the sample standard deviation as your estimate of σ, the standard deviation of that normal distribution. Use a 5% level of significance. Compute degrees of freedom as c–3, where c is the number of test score categories. (The reason df is not computed as c 1 as shown in the chapter is that one additional degree of freedom is lost when the sample mean is used to estimate and one more degree of freedom is lost when the sample standard deviation is used to estimate)

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