Question

complaints about poor service led the department of motor vehicles to investigate wait times at a local branch. during one week, dmv employees randomly asked 100 people leaving this branch how long they had to wait in line. the selected people waited an average of 49.6 minutes, with a standard deviation of 16 minutes. find a 90% confidence interval for the mean wait time at this dmv branch during that week. round your answers to the nearest tenth. <μ<

Explanation:

Step1: Identify the formula

For a large - sample (n = 100) confidence interval of the population mean when the population standard deviation $\sigma$ is unknown (we use the sample standard deviation s instead), the formula is $\bar{x}\pm z_{\alpha/2}\frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z_{\alpha/2}$ is the z - score, s is the sample standard deviation, and n is the sample size.

Step2: Determine the values of $\bar{x}$, s, and n

We are given that $\bar{x}=49.6$, s = 16, and n = 100.

Step3: Find the z - score for a 90% confidence interval

The confidence level is 90%, so $\alpha=1 - 0.90=0.10$ and $\alpha/2=0.05$. Looking up in the standard normal distribution table, $z_{\alpha/2}=z_{0.05}\approx1.645$.

Step4: Calculate the margin of error E

$E = z_{\alpha/2}\frac{s}{\sqrt{n}}=1.645\times\frac{16}{\sqrt{100}}=1.645\times\frac{16}{10}=1.645\times1.6 = 2.632$.

Step5: Calculate the confidence interval

The lower limit is $\bar{x}-E=49.6 - 2.632=46.968\approx47.0$.
The upper limit is $\bar{x}+E=49.6 + 2.632=52.232\approx52.2$.

Answer:

$47.0<\mu<52.2$