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.

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

[tex]1.4-2.447\frac{0.23}{\sqrt{7}}=1.187[/tex]

[tex]1.4+2.447\frac{0.23}{\sqrt{7}}=1.613[/tex]

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

Step-by-step explanation:

Information given

[tex]\bar X =1.4[/tex] represent the sample mean

[tex]\mu[/tex] 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:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)

The degrees of freedom are given by

[tex]df=n-1=7-1=6[/tex]

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

Now replacing we got:

[tex]1.4-2.447\frac{0.23}{\sqrt{7}}=1.187[/tex]

[tex]1.4+2.447\frac{0.23}{\sqrt{7}}=1.613[/tex]

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