I am further interested in knowing if students who missed no classes did better
ID: 3316604 • Letter: I
Question
I am further interested in knowing if students who missed no classes did better on tests than those who missed at least one class. For students who missed no class, the mean of their average test scores was 90.44 and the standard deviation was 10.89. For students who missed at least one class, the mean of their average test was 78.32 and the standard deviation was 18.72. Assume the population variance of average scores for students who missed no class is NOT equal to the population variance of average scores for students who missed at least one class. At the 5% significance level, do we have sufficient evidence to suggest that students who missed no classes did better on the tests than those who missed at least one class? Note that 93 students missed no class and 19 students missed at least one class.
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 = 93
n2= 19
x 1 = 90.44,
x 2 = 78.32,
s1 = 10.89,
s2 = 18.72,
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 = 4.44
DF = 110
t = [ (x1 - x2) - d ] / SE
t = 2.73
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 2.73 We use the t Distribution Calculator to find P(t > 2.73).
Therefore, the P-value in this analysis is 0.004.
Interpret results. Since the P-value (0.004) is less than the significance level (0.05), we have to reject the null hypothesis.
c) At the 5% significance level, we have sufficient evidence to suggest that students who missed no classes did better on the tests than those who missed at least one class.
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