## A textile fiber manufacturer is investigating a new drapery yarn, which the company claims has a mean thread elongation of 12 kilograms with

Question

A textile fiber manufacturer is investigating a new drapery yarn, which the company claims has a mean thread elongation of 12 kilograms with a standard deviation of 0.5 kilograms. The company wishes to test the hypothesis H0: µ = 12 against H1: µ < 12 using a random sample of n = 4 specimens. Calculate the P-value if the observed statistic is Xbar (average) = 11.25. Suppose that the distribution of the sample mean is approximately normal.

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1 year 2021-08-06T15:13:04+00:00 1 Answers 596 views 0

The p-value of the test is 0.0013.

Step-by-step explanation:

The test statistic is:

$$z = \frac{X – \mu}{\frac{\sigma}{\sqrt{n}}}$$

In which X is the sample mean, $$\mu$$ is the value tested at the null hypothesis, $$\sigma$$ is the standard deviation and n is the size of the sample.

12 is tested at the null hypothesis:

This means that $$\mu = 12$$

Standard deviation of 0.5 kilograms.

This means that $$\sigma = 0.5$$

Sample of n = 4 specimens. Observed statistic is Xbar (average) = 11.25.

This means that $$n = 4, X = 11.25$$

Value of the test statistic:

$$z = \frac{X – \mu}{\frac{\sigma}{\sqrt{n}}}$$

$$z = \frac{11.25 – 12}{\frac{0.5}{\sqrt{4}}}$$

$$z = -3$$

P-value:

Probability of finding a sample mean belo 11.25, which is the p-value of z = -3.

Looking at the z-table, z = -3 has a p-value of 0.0013, thus the this is the p-value of the test.