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

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.

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Genesis 2 weeks 2021-10-01T17:10:17+00:00 1 Answer 0

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    2021-10-01T17:12:03+00:00

    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.

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