Question

In a sample of 8 high school students, they spent an average of 28.8 hours each week doing sports with a sample standard deviation of 3.2 hours. Find the 95% confidence interval, assuming the times are normally distributed.

Answer 1

Answer:

$26.12\:<\:\mu\:<\:31.48$

Step-by-step explanation:

Since the population standard deviation $\sigma$ is unknown, and the sample standard deviation $s$, must replace it, the $t$ distribution  must be used for the confidence interval.

The sample size is n=8.

The degree of freedom is $df=n-1$, $\implies df=8-1=7$.

With 95% confidence level, the $\alpha-level$(significance level) is 5%.

Hence with 7 degrees of freedom, $t_{\frac{\alpha}{2} }=2.365$. (Read from the t-distribution table see attachment)

The 95% confidence interval can be found by using the formula:

$\bar X-t_{\frac{\alpha}{2}}(\frac{s}{\sqrt{n} } )\:<\:\mu\:<\:\bar X+t_{\frac{\alpha}{2}}(\frac{s}{\sqrt{n} } )$.

The sample mean is $\bar X=28.8$ hours.

The sample sample standard deviation is $s=3.2$ hours.

We now substitute all these values into the formula to obtain:

$28.8-2.365(\frac{3.2}{\sqrt{8} } )\:<\:\mu\:<\:28.8+2.365(\frac{3.2}{\sqrt{8} } )$.

$26.12\:<\:\mu\:<\:31.48$

We are 95% confident that the population mean is between 26.12 and 31.48  hours.