Using the t-distribution, it is found that since the p-value is greater than 0.01, there is no evidence that the new machine should be rejected at the 0.01 significance level.
At the null hypothesis, it is tested if the average is not less than 25, that is:
[tex]H_0: \mu \geq 25[/tex]
At the alternative hypothesis, it is tested if the average is less than 25, that is:
[tex]H_1: \mu < 25[/tex]
The test statistic is given by:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
The parameters are:
Considering the situation described, the values of the parameters are given as follows:
[tex]\overline{x} = 24.97, \mu = 25, s = 2.16, n = 10[/tex]
Hence the test statistic is:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{24.97 - 25}{\frac{2.16}{\sqrt{10}}}[/tex]
t = -0.04
Using a t-distribution calculator, for a left-tailed test, as we are testing if the mean is less than a value, with 10 - 1 = 9 df and t = -0.04, the p-value is of 0.4844.
Since the p-value is greater than 0.01, there is no evidence that the new machine should be rejected at the 0.01 significance level.
More can be learned about the t-distribution at https://brainly.com/question/13873630
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