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In the morning class, the mean on the first exam was 78 with a standard deviation of 2.3 for

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In the morning class, the mean on the first exam was 78 with a standard deviation of 2.3 for
the 57 students in the class. In the afternoon class, the mean was 81 with a standard deviation
of 5.7 for the 37 students in the class. Determine the 90% confidence interval for the
difference in the means of the two groups
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Maggy Mwangi

Maggy Mwangi Added an answer on Aug 13,2019

Using Minitab Output:

Two-Sample T-Test and CI

Sample   N   Mean StDev SE Mean
1       57 78.00   2.30     0.30
2       37 81.00   5.70     0.94

Difference = μ (1) - μ (2)
Estimate for difference: -3.000
90% CI for difference: (-4.656, -1.344)
T-Test of difference = 0 (vs ≠): T-Value = -3.04 P-Value = 0.004 DF = 43

Maggy Mwangi

Maggy Mwangi Added an answer on Aug 13,2019

When calculating the confidence interval requires calculation of the degrees of freedom and then using the appropriate formula.

Formula for degrees of freedom:

In the morning class, the mean on the first exam was 78 with a standard deviation of 2.3 for

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