Mathematics, 22.07.2021 19:20 lesliemar

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.

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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 z = 1.67 < z_c

, 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

$H_0: p \leq 0.1$

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

H_1: p 0.1

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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 Z_c = 2.327

. 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:

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

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 $\mu = 0.1, \sigma = \sqrt{0.1*0.9}$

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

This means that $n = 100, X = \frac{15}{100} = 0.15$

Value of the test-statistic:

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

$z = \frac{0.15 - 0.1}{\frac{\sqrt{0.1*0.9}}{\sqrt{100}}}$

z = 1.67

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Decision: Critical region

The test statistic is z = 1.67 < z_c , 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

Answer from: Quest

the answers c

Answer from: Quest

80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol