A machine in a factory must be repaired if it produces more than 10% defectives in production. A random sample of 100 items from a day's production contains 15 defectives, and the foreman says that the machine must be repaired. Statistically, does the sample evidence support his decision to repair at the 0.01 significance level? Conduct a test by using both the critical region method and the p-value method.

From the test the person wants, and the sample data, we build the test hypothesis, find the test statistic,  and use this to reach a conclusion both using the critical value and the p-value.

Doing this, the conclusions are:

The test statistic is , meaning that there is not enough evidence to conclude that the proportion of defectives is above 10%, that is, it does not support his decision to repair at the 0.01 significance level.The p-value of the test is 0.0475 > 0.01, meaning that there is not enough evidence to conclude that the proportion of defectives is above 10%, that is, it does not support his decision to repair at the 0.01 significance level.

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Hypothesis:

A machine in a factory must be repaired if it produces more than 10% defectives in production.

At the null hypothesis, we test if it does not have to be repaired, that is,  the proportion is of at most 10%. So

At the alternative hypothesis, we test if it does have to be repaired, that is, the proportion is greater than 10%. So

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Decision rule:

0.01 significance level, using a left-tailed test(testing if the mean is more than a value), which means that:

The critical value is Z with a p-value of 1 - 0.01 = 0.99, so . If the test statistic z is less than this, there is not enough evidence to reject the null hypothesis, that the proportion is of at most 10%, otherwise, there is.The p-value is the probability of finding a sample proportion above the one found. If it is more than 0.01, there is not enough evidence to reject the null hypothesis, otherwise, there is.

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The test statistic is:

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

0.1 is tested at the null hypothesis:

This means that

A random sample of 100 items from a day's production contains 15 defectives.

This means that

Value of the test-statistic:

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Decision: Critical region

The test statistic is , meaning that there is not enough evidence to conclude that the proportion of defectives is above 10%, that is, it does not support his decision to repair at the 0.01 significance level.

Decision: p-value

The p-value of the test is the probability of finding a sample proportion above 0.15, which is 1 subtracted by the p-value of z = 1.67.

Looking at the z-table, z = 1.67 has a p-value of 0.9525.

1 - 0.9525 = 0.0475.

The p-value of the test is 0.0475 > 0.01, meaning that there is not enough evidence to conclude that the proportion of defectives is above 10%, that is, it does not support his decision to repair at the 0.01 significance level.

A similar problem is found at

1 penny on top bottom I think (need pts)