Question

In an effort to estimate the mean of amount spent per customer for dinner at a major Lawrence restaurant, data were collected for a sample of 49 customers. Assume a population standard deviation of $5. If the sample mean is $24.80, what is the 99% confidence interval for the population mean?

Answer 1

Answer:

The 99% confidence interval for the population mean is 22.96 to 26.64

Step-by-step explanation:

Consider the provided information,

A sample of 49 customers. Assume a population standard deviation of $5. If the sample mean is $24.80,

The confidence interval if 99%.

Thus, 1-α=0.99

α=0.01

Now we need to determine $z_{\frac{\alpha}{2}}=z_{0.005}$

Now by using z score table we find that  $z_{\frac{\alpha}{2}}=2.58$

The boundaries of the confidence interval are:

$\mu-z_{\frac{\alpha}{2}}\times \frac{\sigma}{\sqrt{n} }\\24.80-2.58\times \frac{5}{\sqrt{49}}=22.96\\\mu+z_{\frac{\alpha}{2}}\times \frac{\sigma}{\sqrt{n} }\\24.80+2.58\times \frac{5}{\sqrt{49}}=26.64$

Hence, the 99% confidence interval for the population mean is 22.96 to 26.64