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Sagot :
Answer:
The p-value of the test is 0.015 < 0.05, which means that there is sufficient evidence at the 0.05 level to support the company's claim.
Step-by-step explanation:
The company's promotional literature claimed that more than 47% do not fail in the first 1000 hours of their use.
At the null hypothesis, we test if the proportion is of 47% or less, that is:
[tex]H_0: p \leq 0.47[/tex]
At the alternative hypothesis, we test if the proportion is of more than 47%, that is:
[tex]H_1: p > 0.47[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
0.47 is tested at the null hypothesis:
This means that [tex]\mu = 0.47, \sigma = \sqrt{0.47*0.53}[/tex]
A sample of 1300 computer chips revealed that 50% of the chips do not fail in the first 1000 hours of their use.
This means that [tex]n = 1300, X = 0.5[/tex]
Value of the statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.5 - 0.47}{\frac{\sqrt{0.47*0.53}}{\sqrt{1300}}}[/tex]
[tex]z = 2.17[/tex]
P-value of the test and decision:
The p-value of the test is the probability of finding a sample proportion above 0.5, which is 1 subtracted by the p-value of z = 2.17.
Looking at the z-table, z = 2.17 has a p-value of 0.9850
1 - 0.985 = 0.015
The p-value of the test is 0.015 < 0.05, which means that there is sufficient evidence at the 0.05 level to support the company's claim.
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