A survey of 50 retail stores revealed that the average price of a microwave was $375 with a sample standard deviation of $20. Assuming the population is normally distributed,what is the 99% confidence interval to estimate the true cost of the microwave?
A)$367.42 to $382.58
B)$315.00 to $415.00
C)$323.40 to $426.60
D)$335.82 to $414.28

Question

Quizplus · Accepted Answer

The 99% confidence interval can be calculated using the formula for the confidence interval for the mean, which is $\bar{x} \pm z\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z$ is the z-score corresponding to the desired confidence level (for 99%, $z = 2.576$), $\sigma$ is the sample standard deviation, and $n$ is the sample size. Plugging in the given values: $375 \pm 2.576\frac{20}{\sqrt{50}}$ results in the interval $367.42$ to $382.58$.

A Survey of 50 Retail Stores Revealed That the Average