I am interested in knowing if the first test was significantly more difficult fo
ID: 3316602 • Letter: I
Question
I am interested in knowing if the first test was significantly more difficult for students than the second test, since the average score for the first test was less than the average score for the second test. For the first test, the mean score was 87.46 and the standard deviation was 14.69. For the second test, the mean score was 89.13 and the standard deviation was 17.67. Assume that the population variance of scores for the first test is equal to the population variance of scores for the second test. At the 5% significance level, do we have sufficient evidence to suggest that the first test was more difficult for students than the second test ? Note that 112 students took both test.
a) What hypothesis test would be most appropriate to answer this question?
b) Find H0, H1, , n1, n2, x 1, x 2, s1, and s2.
c) Conduct the hypothesis test with the critical value method. Don’t forget to give your conclusion!
Explanation / Answer
Solution:-
= 0.05,
n1 = 112,
n2= 112
x 1 = 87.46,
x 2 = 89.13,
s1 = 14.69,
s2 = 17.67
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1> 2
Alternative hypothesis: 1 < 2
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees offreedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 2.17
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = 222
t = [ (x1 - x2) - d ] / SE
t = - 0.77
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is thesize of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.
The observed difference in sample means produced a t statistic of - 0.77 We use the t Distribution Calculator to find P(t < - 0.77).
Therefore, the P-value in this analysis is 0.221
Interpret results. Since the P-value (0.221) is greater than the significance level (0.05), we cannot reject the null hypothesis.
c) From the above test we do not have sufficient evidence in the favor of the claim that the first test was more difficult for students than the second test.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.