Respuesta :
Answer:
We conclude that the null hypothesis is not rejected and the valve does not performs above the specifications.
Step-by-step explanation:
We are given that the engineer designed the valve such that it would produce a mean pressure of 7.9 pounds/square inch. It is believed that the valve performs above the specifications.
The valve was tested on 16 engines and the mean pressure was 8.2 pounds/square inch with a standard deviation of 0.5.
Let [tex]\mu[/tex] = population mean pressure of the valve
SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \leq[/tex] 7.9 pounds/square inch {means that the valve does not performs above the specifications}
Alternate Hypothesis, [tex]H_a[/tex] : [tex]\mu[/tex] > 7.9 pounds/square inch {means that the valve performs above the specifications}
The test statistics that will be used here is One-sample t test statistics because we don't know about the population standard deviation;
T.S. = [tex]\frac{\bar X -\mu}{{\frac{s}{\sqrt{n} } } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean pressure = 8.2 pounds/square inch
s = sample standard deviation = 0.5
n = sample of engines = 16
So, test statistics = [tex]\frac{8.2-7.9}{{\frac{0.5}{\sqrt{16} } } }[/tex] ~ [tex]t_1_5[/tex]
= 2.40
The decision rule for rejecting the null hypothesis is given by;
- If the test statistics is more than the critical of t at 15 degree of freedom, then we will reject our null hypothesis as it will fall in the rejection region.
- If the test statistics is less than the critical of t at 15 degree of freedom, then we will not reject our null hypothesis as it will not fall in the rejection region.
Now at 0.01 significance level, the t table gives critical value of 2.602 at 15 degree of freedom for one-tailed test. Since our test statistics is less than the critical value of t so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.
Therefore, we conclude that the valve does not performs above the specifications.
