Question

A review session is given in order to find if the group who attends it has a better average than who doesn't. The grades are organized into two independent groups and the data is obtained. The mean and standard deviation of the 29 students who attended the review session are 70.5 and 11.6 percent respectively. The mean and standard deviation of the 14 students who did NOT attend the review session are 64.6 and 16.1 percent respectively. In this problem, you will calculate the 90%, 95% and 99% confidence intervals for the difference in the mean of the two student test groups and determine if there is a true difference in the average of the two groups. We will assume for this part of the problem that we consider the variances are NOT "equal." Perform the following steps:

What is the standard error of the mean for the three confidence intervals?

What is the margin of error for each of the three confidence intervals?

Construct the 90%, 95% and 99% confidence intervals for the "true" difference between the test averages of the two groups, showing first the mean +/- the margin of error, and then showing the range of the interval.

Is there a true "difference" in the means of the review session attenders vs. those who did not? On what information provided by the confidence intervals are you basing your answer?

Answer 1

The formula for estimation is:

μ₁ - μ₂ = (M₁ - M₂) ± ts_{(M1 - M2)}

where:

M₁ & M₂ = sample means
t = t statistic determined by confidence level(90%)
s_{(M1 - M2)} = standard error = √((s²_p/n₁) + (s²_p/n₂))

Calculation

Pooled Variance
s²_p = ((df₁)(s²₁) + (df₂)(s²₂)) / (df₁ + df₂) = 11025.56 / 56 = 196.89

Standard Error
s_{(M1 - M2)} = √((s²_p/n₁) + (s²_p/n₂)) = √((196.89/29) + (196.89/29)) = 3.68

Confidence Interval
μ₁ - μ₂ = (M₁ - M₂) ± ts_{(M1 - M2)} = 5.9 ± (1.67 * 3.68) = 5.9 ± 6.163

90% CI [-0.263, 12.063].

again, 95% confidence interval

Calculation

Pooled Variance
s²_p = ((df₁)(s²₁) + (df₂)(s²₂)) / (df₁ + df₂) = 11025.56 / 56 = 196.89

Standard Error
s_{(M1 - M2)} = √((s²_p/n₁) + (s²_p/n₂)) = √((196.89/29) + (196.89/29)) = 3.68

Confidence Interval
μ₁ - μ₂ = (M₁ - M₂) ± ts_{(M1 - M2)} = 5.9 ± (2 * 3.68) = 5.9 ± 7.382

CI [-1.482, 13.282].

Again 99 %

Calculation

Pooled Variance
s²_p = ((df₁)(s²₁) + (df₂)(s²₂)) / (df₁ + df₂) = 11025.56 / 56 = 196.89

Standard Error
s_{(M1 - M2)} = √((s²_p/n₁) + (s²_p/n₂)) = √((196.89/29) + (196.89/29)) = 3.68

Confidence Interval
μ₁ - μ₂ = (M₁ - M₂) ± ts_{(M1 - M2)} = 5.9 ± (2.67 * 3.68) = 5.9 ± 9.826

99% CI [-3.926, 15.726]

Here if we compate all three Confidence interval it can be asily interpreted that 0 lies in all the three confidence interval hence here the true mean of both the sample can be equal.