Given a test statistic of t=2.315 of a left tailed test with n=8

Assuming this complete question :"Given a test statistic of t=2.315 of a left-tailed test with n=8, use a 0.05 significance level to test a claim that the mean of a given population is equal to 110.

Find the range of values for the P-value and state the initial conclusion 1 point) 0.05<P-value <0.10; reject the null hypothesis

0.05<P-value <0.10, fail to reject the null hypothesis

0.025 < P-value <0.05; reject the null hypothesis

0.025< P-value<0.05; fail to reject the null hypothesis "

For this case they want to test if the population mean is 110 or no, the systemof hypothesis are:

Null hypothesis: [tex] \mu =110[/tex]

Alternative hypothesis:[tex]\mu \neq 110[/tex]

The sample size on this case is n=8, then the degrees of freedom are given by:

[tex] df = n-1= 8-1=7[/tex]

The statistic is given by:

[tex] t= \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}[/tex]

For this case the value of the statistic is given [tex] t = 2.315[/tex]

Since we are using a bilateral test the p value would be given by:

[tex] p_v = 2*P(t_{7}>2.315) =0.054[/tex]

And we can use the following excel code to find it:

"=2*(1-T.DIST(2.315;7;TRUE))"

Since the p value is higher than the significance level given we FAIL to reject the null hypothesis. And the best conclusion would be:

0.05<P-value <0.10, fail to reject the null hypothesis