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Akimi4 [234]
3 years ago
13

An automobile engineer is revising a design for a conical chamber that was originally specified to be 12 inches long with a circ

ular base diameter of 5.7 inches. In the new design, the chamber is scaled by a factor of 1.5. What is the volume of the revised chamber? Round your answer to two decimal places.
Mathematics
1 answer:
icang [17]3 years ago
4 0

The scale factor is 1.5, which means that the revised chamber is larger. The new dimensions are just the old dimensions times 1.5.

The volume of the revise chamber is:

V = (1/3) (PI) [(D/2)^2] (H)

= (1/3) (PI) {[(5.7)(1.5)/2]^2} [(12)(1.5)]

= 344.4874 in3

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\chi^2 =\frac{22-1}{259.21} 166.41 =13.482

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In order to find the p value we can use the following code in excel:

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If we compare the p value and the significance level provided we see that p_v >\alpha so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population deviation is not significantly lower than 16.1 at 1% of significance.

Step-by-step explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

n=22 represent the sample size

\alpha=0.01 represent the confidence level  

s^2 =12.9^2=166.41 represent the sample variance obtained

\sigma^2_0 =16.1^2 =259.21 represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population deviation is smaller than 16.1 (that's equivalent to check if the population variance is lower than 259.21:

Null Hypothesis: \sigma^2 \geq 259.21

Alternative hypothesis: \sigma^2

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For this test we can use the following statistic:

\chi^2 =\frac{n-1}{\sigma^2_0} s^2

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

\chi^2 =\frac{22-1}{259.21} 166.41 =13.482

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 21. And since is a left tailed test the p value would be given by:

p_v =P(\chi^2

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(13.482,21,TRUE)"

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If we compare the p value and the significance level provided we see that p_v >\alpha so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population deviation is not significantly lower than 16.1 at 1% of significance.

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