The valve was tested on 270 engines and the mean pressure was 6.6 lbs/square inch. Assume the variance is known to be 0.49. If the valve was designed to produce a mean pressure of 6.5 lbs/square inch, is there sufficient evidence at the 0.1 level that the valve does not perform to the specifications

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Answer:

[tex]z=\frac{6.6-6.5}{\frac{0.7}{\sqrt{270}}}=2.347[/tex]      

The p value for this case would be given by"

[tex]p_v =2*P(z>2.347)=0.0189[/tex]  

For this case since the p value is higher than the significance level we don't have enough evidence to conclude that the true mean is significantly different from 6.5 lbs/square inch at 10% of significance. So then there is not enough evidence to conclude that the valve does not perform to the specifications

Step-by-step explanation:

Information given

[tex]\bar X=6.6[/tex] represent the sample mean

[tex]s=\sqrt{0.49}= 0.7[/tex] represent the population deviation

[tex]n=270[/tex] sample size      

[tex]\mu_o =6.5[/tex] represent the value that we want to test    

[tex]\alpha=0.1[/tex] represent the significance level

z would represent the statistic

[tex]p_v[/tex] represent the p value for the test

Hypothesis to verify

We want to verify if the true mean for this case is equal to 6.5 lbs/square inch or not , the system of hypothesis would be:      

Null hypothesis:[tex]\mu= 6.5[/tex]      

Alternative hypothesis:[tex]\mu \neq 6.5[/tex]      

The statistic for this case is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)      

And replacing we got:

[tex]z=\frac{6.6-6.5}{\frac{0.7}{\sqrt{270}}}=2.347[/tex]      

The p value for this case would be given by"

[tex]p_v =2*P(z>2.347)=0.0189[/tex]  

For this case since the p value is higher than the significance level we don't have enough evidence to conclude that the true mean is significantly different from 6.5 lbs/square inch at 10% of significance. So then there is not enough evidence to conclude that the valve does not perform to the specifications

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