The following data represent a random sample the golf scores for a particular go

Question

The following data represent a random sample the golf scores for a particular golfer from a course that he regularly plays. This golfer would like to test it his golf scores at this course follow the normal probability distribution with mu = 85.0 and sigma = 5.0 using alpha = 0.05. What is the critical value for this hypothesis test ? Round to three decimal places. 84 82 72 86 83 82 89 80 94 84 85 85 89 87 92 88 94 87 86 84 89 84 87 89 96 72 88 89 79 90 76 86 87 89 96 A. 9.448 B. 7.815 C. 11.070 D. 5.991

Explanation / Answer

The Chi-square goodness of fit test is one of the appropriate tests to see if the data comes from normal distribution. For Chi-square goodness of fit test to employ in this situation, one should compute the interval of scores and corresponding frequencies. The degrees of freedom is (interval-1).

The frequency distribution is as follows:

Look for Chi-square table. Therefore, for 4-1=3 degrees of freedom and alpha=0.05, the critical Chi-sqaure value is 7.815. Answer: B.

Interval Frequency 72-79 4 80-87 17 88-95 12 96-103 2

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