A sample of 32 orders for a home shopping channel has an average of $78.25.The standard deviation for all orders (the population)is $37.50.A 90% confidence interval around this sample mean would be:
A)$\$ 78.25 \pm \$ 37.50$
B)$\$ 78.25 \pm \$ 10.87$
C)$\$ 78.25 \pm \$ 6.63$
D)$\$ 37.50 \pm \$ 6.63$
E) $\$ 37.50 \pm \$ 1.64$

Question

Quizplus · Accepted Answer

To calculate a 90% confidence interval for the sample mean, we use the formula: $$ \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) $$ where $\bar{x}$ is the sample mean, $z$ is the z-score corresponding to the desired confidence level (for 90% confidence, $z \approx 1.645$), $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values: $$ 78.25 \pm 1.645 \left( \frac{37.50}{\sqrt{32}} \right) \approx 78.25 \pm 10.87 $$

A Sample of 32 Orders for a Home Shopping Channel $78.25±$37.50\$ 78.25 \pm \$ 37.50$78.25±$37.50

A Sample of 32 Orders for a Home Shopping Channel $\$ 78.25 \pm \$ 37.50$