Question

the average selling price of a smartphone purchased by a random sample of 41 customers was $303. assume the population standard deviation was $33.
a. construct a 90% confidence interval to estimate the average selling price in the population with this sample.
b. what is the margin of error for this interval?
a. the 90% confidence interval has a lower limit of $ and an upper limit of $ (round to the nearest cent as needed.)

Explanation:

Step1: Identify the formula for confidence interval

The formula for a confidence interval for the population mean when the population standard - deviation $\sigma$ is known is $\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z_{\alpha/2}$ is the z - score, $\sigma$ is the population standard deviation, and $n$ is the sample size.
For a 90% confidence interval, $\alpha = 1 - 0.90=0.10$, and $\alpha/2 = 0.05$. The $z$ - score $z_{\alpha/2}=z_{0.05}\approx1.645$. Given $\bar{x} = 303$, $\sigma = 33$, and $n = 41$.

Step2: Calculate the margin of error

The margin of error $E=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$.
$E = 1.645\times\frac{33}{\sqrt{41}}\approx1.645\times\frac{33}{6.4031}\approx1.645\times5.1538\approx8.48$.

Step3: Calculate the lower and upper limits of the confidence interval

The lower limit $LL=\bar{x}-E=303 - 8.48 = 294.52$.
The upper limit $UL=\bar{x}+E=303 + 8.48 = 311.48$.

Answer:

a. The 90% confidence interval has a lower limit of $\$294.52$ and an upper limit of $\$311.48$.
b. The margin of error for this interval is $\$8.48$.