Respuesta :
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 [tex]z = 1.67 < z_c[/tex], 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
[tex]H_0: p \leq 0.1[/tex]
At the alternative hypothesis, we test if it does have to be repaired, that is, the proportion is greater than 10%. So
[tex]H_1: p > 0.1[/tex]
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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 [tex]Z_c = 2.327[/tex]. 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:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
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 [tex]\mu = 0.1, \sigma = \sqrt{0.1*0.9}[/tex]
A random sample of 100 items from a day's production contains 15 defectives.
This means that [tex]n = 100, X = \frac{15}{100} = 0.15[/tex]
Value of the test-statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.15 - 0.1}{\frac{\sqrt{0.1*0.9}}{\sqrt{100}}}[/tex]
[tex]z = 1.67[/tex]
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Decision: Critical region
The test statistic is [tex]z = 1.67 < z_c[/tex], 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 https://brainly.com/question/24326664
