In an examination of purchasing patterns of shoppers, a sample of 16 shoppers revealed that they spent, on average, $54 per hour of shopping. Based on previous years, the population standard deviation is thought to be $21 per hour of shopping. Assuming that the amount spent per hour of shopping is normally distributed, find a 90% confidence interval for the mean amount.
A) [$45.36, $62.64]
B) [$47.27, $60.73]
C) [$51.84, $56.16]
D) [$52.32, $55.68]

Question

Quizplus · Accepted Answer

The 90% confidence interval for the mean amount spent per hour of shopping can be calculated using the formula for a confidence interval when the population standard deviation is known: $\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $Z$ is the Z-value corresponding to the desired confidence level (for 90%, $Z = 1.645$), $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values: $54 \pm 1.645 \frac{21}{\sqrt{16}}$ results in the interval $[45.36, 62.64]$.

In an Examination of Purchasing Patterns of Shoppers, a Sample