Question

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.

Answer 1

Answer:

$t = \frac{18-15}{\frac{6}{\sqrt{16}}}= 2$

$df = n-1= 16-1=15$

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

$p_v= P(t_{15}>2) = 0.0639$

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:

$n = 16$ represent the sample size

$\bar X = 18$ represent the sample mean for the drying time

$s= 6$ 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: $\mu \leq 15$

Alternative hypothesis: $\mu >15$

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

$t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}$

And replacing we have:

$t = \frac{18-15}{\frac{6}{\sqrt{16}}}= 2$

Now we can find the degrees of freedom given by:

$df = n-1= 16-1=15$

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

$p_v= P(t_{15}>2) = 0.0639$

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.