The management of a department store is interested in estimating the difference between the mean credit purchases of customers using the store's credit card versus those customers using a national major credit card.You are given the following information. $$\begin{array} { l l l }
& \text { Store's Card } & \text { Major Credit Card } \
\text { Sample size } & 64 & 49 \
\text { Sample mean } & \$ 140 & \$ 125 \
\text { Population standard deviation } & \$ 10 & \$ 8
\end{array}$$ At 95% confidence, the margin of error is
A) 1.694.
B) 3.32.
C) 1.96.
D) 15.

Question

The management of a department store is interested in estimating the difference between the mean credit purchases of customers using the store's credit card versus those customers using a national major credit card.You are given the following information. $$\begin{array} { l l l } 
& \text { Store's Card } & \text { Major Credit Card } \
\text { Sample size } & 64 & 49 \
\text { Sample mean } & \$ 140 & \$ 125 \
\text { Population standard deviation } & \$ 10 & \$ 8
\end{array}$$  At 95% confidence, the margin of error is
A) 1.694.
B) 3.32.
C) 1.96.
D) 15.

Quizplus · Accepted Answer

The margin of error for the difference between two means is calculated using the formula: $ ME = Z \times \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} $. For a 95% confidence level, Z is approximately 1.96. Substituting the given values: $ ME = 1.96 \times \sqrt{\frac{10^2}{64} + \frac{8^2}{49}} = 3.32 $.

The Management of a Department Store Is Interested in Estimating