Question

An engineer designed a valve that will regulate water pressure on an automobile engine. The engineer designed the valve such that it would produce a mean pressure of 7.97.9 pounds/square inch. It is believed that the valve performs above the specifications. The valve was tested on 1616 engines and the mean pressure was 8.28.2 pounds/square inch with a standard deviation of 0.50.5. A level of significance of 0.010.01 will be used. Assume the population distribution is approximately normal. Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.

Answer 1

We conclude that the null hypothesis is not rejected and the valve does not performs above the specifications.

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
`\mu` = population mean pressure of the valve

SO, Null Hypothesis,
`H_0` :
`\mu \leq` 7.9 pounds/square inch {means that the valve does not performs above the specifications}

Alternate Hypothesis,
`H_a` :
`\mu` > 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. =
`\frac{\bar X -\mu}{{(s)/(√(n) ) } }` ~
`t_n_-_1`

where,
`\bar X` = sample mean pressure = 8.2 pounds/square inch

s = sample standard deviation = 0.5

n = sample of engines = 16

So, test statistics =
`\frac{8.2-7.9}{{(0.5)/(√(16) ) } }` ~
`t_1_5`

= 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.