The frequency distribution shows the results of 200 test scores. Are the test sc

Question

The frequency distribution shows the results of 200 test scores. Are the test scores normally distributed?

PART C. Calculate the test statistic

PART D. Decide whether to reject or fail to reject the null hypothesis

The frequency distribution shows the results of 200 test scores. Are the test scores normally distributed? Use = 0.01. Complete parts (a) through (d) Class boundaries Frequency, f 49.5-58.5 20 58.5-67.5 62 67.5-76.5 79 76.5-85.5 35 85.5-94.5 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 normality, the null and alternative hypotheses are as follows Ho: The test scores have a normal distribution Ha: The test scores do not have a normal distribution a. Find the expected frequencies lass boundaries equency, f xpected frequency 49.5-58.5 20 58.5-67.5 62 67.5-76.5 79 76.5-85.5 35 85.5-94.5 (Round to the nearest integer as needed.)

Explanation / Answer

f=(20,62,79,35,4), where f is frequency
o=(54,63,72,81,90) , where o is mid value of class boundary
mean=sum(f.o)/sum of frequency f

mean= 69.345
e=(69.345,69.345,69.345,69.345,69.345), where e is expected frequency
(o-e)^2= 235.469025 40.259025   7.049025 135.839025 426.629025

(o-2)^2/e= 3.3956165 0.5805613 0.1016515 1.9588871 6.1522680

sum of (o-2)^2/e= calculated chi square=12.18898

tabulated chi square with 4 degree of freedom and alpha=.01=15.086

(a)expected frequency= e=(69.345,69.345,69.345,69.345,69.345)

(b) critical value is greater than 15.086.

(c) calculated test statistics is 12.18898.

(d) calulated value is less than tabulated value. so do not rejected null hypothesis.

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