Question

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A researcher collected a simple random sample of the cents portions from 100 checks and from 100 credit card charges. The cents portions of the checks have a mean of 23.8 cents and a standard deviation of 32.0 cents. The cents portions of the credit charges have a mean of 47.6 cents and a standard deviation of 33.5 cents. Construct a 95% confidence interval for the mean difference between the cent portions of credit cards and checks.

Answer 1

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1)

Pooled Variance
s²_p = ((df₁)(s²₁) + (df₂)(s²₂)) / (df₁ + df₂) = 212478.75 / 198 = 1073.12

Standard Error
s_{(M1 - M2)} = √((s²_p/n₁) + (s²_p/n₂)) = √((1073.12/100) + (1073.12/100)) = 4.63

Confidence Interval
μ₁ - μ₂ = (M₁ - M₂) ± ts_{(M1 - M2)} = 23.8 ± (1.97 * 4.63) = 23.8 ± 9.136

μ₁ - μ₂ = (M₁ - M₂) = 23.8, 95% CI [14.664, 32.936].