Mathematics, 04.12.2019 04:31, payshencec21

An engineer has designed a valve that will regulate water pressure on an automobile engine. the valve was tested on 100100 engines and the mean pressure was 7.27.2 lbs/square inch. assume the variance is known to be 0.810.81. if the valve was designed to produce a mean pressure of 7.37.3 lbs/square inch, is there sufficient evidence at the 0.010.01 level that the valve does not perform to the specifications? state the null and alternative hypotheses for the above scenario.

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answered: tiggyandrep2dbee

Null hypothesis: $\mu \geq 7.3$

Alternative hypothesis: $\mu < 7.3$

$z=\frac{7.2-7.3}{\frac{0.81}{\sqrt{100}}}=-1.235$

p_v =P(z

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so then we can conclude that the mean pressure for the automobile engines is not significantly less than 7.3 at 1% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X=7.2$ represent the sample mean

$\sigma=0.81$ represent the population standard deviation

n=100 sample size

$\mu_o =7.3$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean pressure is lower than the specificated value 7.3, the system of hypothesis are :

Null hypothesis: $\mu \geq 7.3$

Alternative hypothesis: $\mu < 7.3$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{7.2-7.3}{\frac{0.81}{\sqrt{100}}}=-1.235$

4) P-value

Since is a one-side lower test the p value would given by:

p_v =P(z

5) Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean pressure for the automobile engines is not significantly less than 7.3 at 1% of significance.

answered: Guest

take the top number and divide it by a the bottom number

step-by-step explanation:

1/2 1 divied by 2 = .5

answered: Guest

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