Question

A manufacturer of widgets wants to test a new widget-producing machine to determine if it can make an average of 25 widgets per second before deciding to invest in the machine. The standard to reject the new machine is if it makes an average of less than 25 widgets per second. Here are data from a small random sample:
25.6, 26.2, 22.5, 20.5, 26.4, 27.4, 23.6, 26.9, 25.7, 24.9

Assuming the population follows a normal distribution, is there evidence that the new machine should be rejected at the 0.01 significance level? State the hypotheses, list and check the conditions, calculate the test statistic, find the p-value, and make a conclusion in a complete sentence related to the scenario.

Answer 1

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.

What are the hypothesis tested?

At the null hypothesis, it is tested if the average is not less than 25, that is:

$H_0: \mu \geq 25$

At the alternative hypothesis, it is tested if the average is less than 25, that is:

$H_1: \mu < 25$

What is the test statistic?

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

• $\overline{x}$ is the sample mean.

• $\mu$ is the value tested at the null hypothesis.

• s is the standard deviation of the sample.

• n is the sample size.

Considering the situation described, the values of the parameters are given as follows:

$\overline{x} = 24.97, \mu = 25, s = 2.16, n = 10$

Hence the test statistic is:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{24.97 - 25}{\frac{2.16}{\sqrt{10}}}$

t = -0.04

What is the p-value and the conclusion?

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 brainly.com/question/13873630

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