Respuesta :
Answer:
The p-value of the test is 0.004 < 0.02, which means that there is sufficient evidence at the 0.02 level to support the company's claim.
Step-by-step explanation:
The company's promotional literature claimed that over 32% do not fail in the first 1000 hours of their use.
At the null hypothesis, we test if the proportion is of at most 32%, that is:
[tex]H_0: p \leq 0.32[/tex]
At the alternative hypothesis, we test if the proportion is more than 32%, that is:
[tex]H_1: p > 0.32[/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.
0.32 is tested at the null hypothesis:
This means that [tex]\mu = 0.32 \sigma = \sqrt{0.32*0.68}[/tex]
A sample of 1700 computer chips revealed that 35% of the chips do not fail in the first 1000 hours of their use.
This means that [tex]n = 1700, X = 0.35[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.35 - 0.32}{\frac{\sqrt{0.32*0.68}}{\sqrt{1700}}}[/tex]
[tex]z = 2.65[/tex]
P-value of the test and decision:
The p-value of the test is the probability of finding a sample proportion above 0.35, which is 1 subtracted by the p-value of z = 2.65.
Looking at the z-table, z = 2.65 has a p-value of 0.9960.
1 - 0.9960 = 0.004.
The p-value of the test is 0.004 < 0.02, which means that there is sufficient evidence at the 0.02 level to support the company's claim.
