Question

In a random sample of 7 residents of the state of Maine, the mean waste recycled per person per day was 1.4 pounds with a standard deviation of 0.23 pounds. Determine the 95% confidence interval for the mean waste recycled per person per day for the population of Maine. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answer 1

`df=n-1=7-1=6`

The Confidence desired is 0.95 or 95%, then the significance is
`\alpha=0.05` and
`\alpha/2 =0.025`, and the critical value in the t distribution with 6 degrees of freedom would be
`t_(\alpha/2)=2.447`

`1.4-2.447(0.23)/(√(7))=1.187`

`1.4+2.447(0.23)/(√(7))=1.613`

The 95% confidence interval for the true mean would be between (1.187;1.613)

Explanation:

Information given

`\bar X =1.4` represent the sample mean

`\mu` population mean (variable of interest)

s=0.23 represent the sample standard deviation

n=7 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

The degrees of freedom are given by

`df=n-1=7-1=6`

The Confidence desired is 0.95 or 95%, then the significance is
`\alpha=0.05` and
`\alpha/2 =0.025`, and the critical value in the t distribution with 6 degrees of freedom would be
`t_(\alpha/2)=2.447`

Now replacing we got:

`1.4-2.447(0.23)/(√(7))=1.187`

`1.4+2.447(0.23)/(√(7))=1.613`

The 95% confidence interval for the true mean would be between (1.187;1.613)