An engineer has designed a valve that will regulate water pressure on an automobile engine. The valve was tested on 150 engines and the mean pressure was 6.3 pounds/square inch (psi). Assume the population variance is 0.49. If the valve was designed to produce a mean pressure of 6.2 psi, is there sufficient evidence at the 0.05 level that the valve does not perform to the specifications?

We are given that The valve was tested on 150 engines and the mean pressure was 7.7 lbs/square inch.

We are also given that the valve was designed to produce a mean pressure of 7.6 lbs/square inch

So,  

Null hypothesis:  

Alternate hypothesis :  

Since n > 30 and population standard deviation is given

So, We will use z test

Formula :  

Substitute the values

refer the z table for p value

so, p value is 0.9927

Since it is a two tailed test So, p = 2(1-  0.9927) = 0.0146

α = 0.1

p value< α

So, we reject null hypothesis

Hence There is  sufficient evidence at the 0.1 level that the valve does not perform to the specifications

Step-by-step explanation:

No, there is insufficient evidence at the 0.02 level that the valve does not perform to the specifications.

Step-by-step explanation:

We are given that an engineer has designed a valve that will regulate water pressure on an automobile engine. The valve was tested on 170 engines and the mean pressure was 7.5 pounds/square inch (psi). Assume the population variance is 0.36 and the valve was designed to produce a mean pressure of 7.4 psi.

We have to test if there sufficient evidence at the 0.02 level that the valve does not perform to the specifications.

Let, NULL HYPOTHESIS, H_0H

0

: \muμ = 7.4 psi {means that the valve perform to the specifications}

ALTERNATE HYPOTHESIS, H_1H

1

: \mu\neqμ



= 7.4 psi {means that the valve does not perform to the specifications}

The test statistics that will be used here is One-sample z-test;

T.S. = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }

n

σ

X

ˉ

−μ

~ N(0,1)

where, \bar X

X

ˉ

= sample mean pressure = 7.5 psi

\sigmaσ = population standard deviation = \sqrt{Variance}

Variance

= \sqrt{0.36}

0.36

= 0.6

n = sample of engines = 170

So, test statistics = \frac{7.5-7.4}{\frac{0.6}{\sqrt{170} } }

170

0.6

7.5−7.4

= 2.173

Now, at 0.02 significance level z table gives critical value of 2.3263. Since our test statistics is less than the critical value of z so we have insufficient evidence to reject null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the valve perform to the specifications and there is not sufficient evidence at the 0.02 level that the valve does not perform to the specifications.

step-by-step explanation:

6.48/4 4 goes into 6.48 1.62 times