a. To determine Explain whether the chi square test for goodness of fit is statistically significant by using the 0.01 significance level. Explanation Step 1: State the null and research hypotheses. Null hypothesis: There is no difference between the observed and expected counts. Research hypothesis: There is significant difference between the observed and expected counts. Step 2: Comparison distribution and degrees of freedom. The chi square test for goodness of fit is used for the comparison. The degrees of freedom can be obtained as follows: df = N c a t e g o r i e s − 1 = 6 − 1 = 5 Step 3: Cut off sample scores From the Table A-4: Cutoff scores for the chi square distribution, the value corresponding to 0.01 level at 5 degrees of freedom is 15.087. Step 4: Chi square test statistic value. The chi square test statistic value is obtained by following the below steps: Determine the observed frequencies by observing the sample data from the information given. From the information given, the actual observed frequencies are 5, 15, 90, 120, 50 and 20. Determine the expected frequency from each category of the given data. The expected number for a category is obtained by using the formula: E = N × percent value 100 . Here, the total number of observations is calculated below: N = 5 + 15 + 90 + 120 + 50 + 20 = 300. Using the given information on expected percentages, the calculation for the expected numbers can be done as follows: Category Expected percentage Expected count 1 2% 6 ( = 300 × 2 100 ) 2 14% 42 ( b. To determine Explain whether the chi square test for goodness of fit is statistically significant by using the 0.01 significance level.

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Chapter 13, Problem 13PP

To determine

Explain whether the chi square test for goodness of fit is statistically significant by using the 0.01 significance level.

To determine

Explain whether the chi square test for goodness of fit is statistically significant by using the 0.01 significance level.

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