Question

a survey of 33 adults found that the mean age of a persons primary vehicle is 5.6 years. assuming the standard deviation of the population is 0.8 year; find the 99% confidence interval of the population mean. is this a z - interval or a t - interval?

Explanation:

Step1: Determine the critical value

For a 99% confidence interval, the significance level $\alpha = 1 - 0.99=0.01$. Then $\frac{\alpha}{2}=0.005$. Looking up in the standard normal distribution table (or using a calculator), the critical value $z_{\frac{\alpha}{2}}=z_{0.005} = 2.576$.

Step2: Calculate the margin of error

The formula for the margin of error $E$ is $E = z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$, where $\sigma = 0.8$ (population standard deviation) and $n = 33$ (sample size).
Substitute the values: $E=2.576\times\frac{0.8}{\sqrt{33}}\approx2.576\times\frac{0.8}{5.745}\approx2.576\times0.139\approx0.36$.

Step3: Calculate the confidence interval

The confidence interval for the population mean $\mu$ is given by $\bar{x}-E<\mu <\bar{x} + E$, where $\bar{x}=5.6$ (sample mean).
Substitute the values: $5.6 - 0.36<\mu<5.6 + 0.36$, so $5.24<\mu<5.96$.

Since the population standard deviation $\sigma$ is known, this is a z - interval.

Answer:

The 99% confidence interval for the population mean is $(5.24,5.96)$ and it is a z - interval.