Solution:-
a)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: u1 = u 2
Alternative hypothesis: u1 u 2
Note that these hypotheses constitute a two-tailed test.
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 of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) +
(s22/n2)]
SE = 1.39414
DF = 6
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 the size 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 the population means, and SE is the standard error.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 6 degrees of freedom is more extreme than -2.73; that is, less than -2.73 or greater than 2.73.
Thus, the P-value = 0.034
Interpret results. Since the P-value (0.034) is less than the significance level (0.05), we cannot accept the null hypothesis.
From the above test we do not have sufficient evidence in the favor of the claim that u1 = u 2.
b)
95% confidence interval for the difference in average final exam grades is C.I = ( 0.389, 7.211).
C.I = 3.80 + 2.447*1.39414
C.I = 3.80 + 3.41146
C.I = ( 0.389, 7.211)
c) Effect size is 0.9357.
d = 0.9357
This is the first homework assignment for STA 3112. It is due in class on Monday,...
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