The manufacturers of a deodorant claim that the mean drying time of their product is, at most, 15 minutes. A sample consisting of 16 cans of the product was used to test the manufacturer's claim. The experiment yielded a mean drying time of 18 minutes with a standard deviation of 6 minutes. Find the t-test and the p-value, and state your conclusion at the 5% significance level.
O t = 2, 2.5% < p-value < 5%, reject the null hypothesis. There enough evidence to conclude that the mean drying time is greater than 15 minutes.
O t = 2, 2.5% < p-value < 5%, fail to reject the null hypothesis. There is not enough evidence to conclude that the mean drying time is greater than 15 minutes.
O t = 2, p-value > 5%, reject the null hypothesis. There enough evidence to conclude that the mean drying time is greater than 15 minutes.
O t = 2, p-value > 5%, fail to reject the null hypothesis. There is not enough evidence to conclude that the mean drying time is greater than 15 minutes.

And for this case the p value is higher than the significance level of 0.05 so then we have enough evidence to FAIL to reject the null hypothesis and the best conclusion for this case is:

t = 2, p-value > 5%, fail to reject the null hypothesis. There is not enough evidence to conclude that the mean drying time is greater than 15 minutes.

Step-by-step explanation:

For this problem we have the following info given:

[tex]n = 16[/tex] represent the sample size

[tex]\bar X = 18[/tex] represent the sample mean for the drying time

[tex] s= 6[/tex] represent the sample deviation

We want to test the claim that the mean drying time of their product is, at most, 15 minutes, so then the system of hypothesis are:

Null hypothesis: [tex]\mu \leq 15[/tex]

Alternative hypothesis: [tex]\mu >15[/tex]

The statistic is given by this formula since we don't know the population deviation:

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

And replacing we have:

[tex] t = \frac{18-15}{\frac{6}{\sqrt{16}}}= 2[/tex]

Now we can find the degrees of freedom given by:

[tex] df = n-1= 16-1=15[/tex]

And the p value since we have a right tailed test is given by:

[tex] p_v= P(t_{15}>2) = 0.0639[/tex]

And for this case the p value is higher than the significance level of 0.05 so then we have enough evidence to FAIL to reject the null hypothesis and the best conclusion for this case is:

t = 2, p-value > 5%, fail to reject the null hypothesis. There is not enough evidence to conclude that the mean drying time is greater than 15 minutes.