Question

A pharmaceutical company is testing a new drug to increase memorization ability. It takes a sample of individuals and splits them randomly into two groups. After the drug regimen is completed, all members of the study are given a test for memorization ability with higher scores representing a better ability to memorize. Those 23 participants on the drug had an average test score of 31.622 (SD = 4.794) while those 21 participants not on the drug had an average score of 32.04 (SD = 5.335). You use this information to create a 95% confidence interval for the difference in average test score. What is the margin of error? Assume the population standard deviations are equal.

Answer 1

Since the population standard deviations are equal hence pooled standard deviation is used as:

The formula for estimation of confidence interval is:

μ₁ - μ₂ = (M₁ - M₂) ± ts_{(M1 - M2)}

where:

M₁ & M₂ = sample means
t = t statistic determined by confidence level and degree of freedom, n-1 by T table shown below
s_{(M1 - M2)} = standard error = √((s²_p/n₁) + (s²_p/n₂))

Pooled Variance
s²_p = ((df₁)(s²₁) + (df₂)(s²₂)) / (df₁ + df₂) = 1074.86 / 42 = 25.59

Standard Error
s_{(M1 - M2)} = √((s²_p/n₁) + (s²_p/n₂)) = √((25.59/23) + (25.59/21)) = 1.53

Confidence Interval
μ₁ - μ₂ = (M₁ - M₂) ± ts_{(M1 - M2)} = 0.418 ± (2.02 * 1.53) = 0.418 ± 3.08136

Hence 95% confidenc interval is

95% CI [-2.66336, 3.49936].

and Margin of error is t* s_{(M1 - M2)}

=(2.02 * 1.53)

=3.08136

The T table used is: