Question

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

​PART B. Determine the critical value and the rejected region

The frequency distribution shows the results of 200 test scores. Are the test scores normally distributed? Use α= 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.)

PART C. Calculate the test statistic

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

Answer 1

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.