1 answer:
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Answer:
The 99% of a confidence interval for the average maximum HP for the experimental engine.
(536.46, 603.54)
Step-by-step explanation:
<u><em>Step:-1</em></u>
Given that the mean of the Population = 540HP
Given that the size of the sample 'n' = 9
Given that the mean of the sample = 570HP
Given that the sample standard deviation = 30HP
<u><em>Step(ii):-</em></u>
<u><em>Degrees of freedom = n-1 =9-1 =8</em></u>
<u><em>t₀.₀₀₅ = 3.3554</em></u>
The 99% of a confidence interval for the average maximum HP for the experimental engine.
(570 - 33.54 , 570+33.54)
(536.46 , 603.54)
<u><em>Final answer</em></u> :-
<em>The 99% of a confidence interval for the average maximum HP for the experimental engine.</em>
<em>(536.46, 603.54)</em>
<em></em>
Answer:
The p-value of the test is 0.0853 > 0.05, which means that there is not enough evidence to reject the manufacturer's claim based on this observation.
Step-by-step explanation:
A manufacturer of nails claims that only 4% of its nails are defective.
At the null hypothesis, we test if the proportion is of 4%, that is:
At the alternative hypothesis, we test if the proportion is more than 4%, that is:
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.
4% is tested at the null hypothesis
This means that
A random sample of 20 nails is selected, and it is found that two of them, 10%, are defective.
This means that
Value of the test statistic:
P-value of the test and decision:
Considering an standard significance level of 0.05.
The p-value of the test is the probability of finding a sample proportion above 0.1, which is 1 subtracted by the p-value of z = 1.37.
Looking at the z-table, z = 1.37 has a p-value of 0.9147
1 - 0.9147 = 0.0853
The p-value of the test is 0.0853 > 0.05, which means that there is not enough evidence to reject the manufacturer's claim based on this observation.