Answer to Question #339430 in Statistics and Probability for Liam

Question

survey of 31 randomly selected students finds that they save a mean of $82 per semester by

using a website. Assume the date comes from a normal distribution and the sample standard deviation is $18 per month.

Confidence Interval: What is the 99% confidence interval to estimate the population mean? (Round your

answers to two decimal places.)

____< u < _____

Answer 1

The critical value for $\alpha = 0.01$ and $df = n-1 = 30$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.749996.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}},\bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(82- 2.749996\times\dfrac{18}{\sqrt{31}},$$

$$82+ 2.749996\times\dfrac{18}{\sqrt{31}})$$

$$=(73.11, 90.89)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $73.11 < \mu < 90.89 ,$ which indicates that we are 99% confident that the true population mean $\mu$  is contained by the interval $(73.11, 90.89).$

$$73.11 < \mu < 90.89$$