Question

The piston diameter of a certain hand pump is 0.5 inch. The manager determines that the diameters are normally distributed, with a mean of 0.5 inch and a standard deviation of 0.006 inch. After recalibrating the production machine, the manager randomly selects 22 pistons and determines that the standard deviation is 0.0044 inch. Is there significant evidence for the manager to conclude that the standard deviation has decreased at the $\alpha = 0.10$ level of significance? What are the correct hypotheses for this test? The null hypothesis is $H_0: \sigma = 0.006$. The alternative hypothesis is $H_1: \sigma < 0.006$. Calculate the value of the test statistic $\chi^2 = 11.293$ (Round to three decimal places as needed.) Use technology to determine the P-value for the test statistic The P-value is (Round to three decimal places as needed.)

          The piston diameter of a certain hand pump is 0.5 inch. The manager determines that the diameters are normally distributed, with a mean of 0.5 inch and a standard deviation of 0.006 inch. After recalibrating the production machine, the manager randomly selects 22 pistons and determines that the standard deviation is 0.0044 inch. Is there significant evidence for the manager to conclude that the standard deviation has decreased at the $\alpha = 0.10$ level of significance?

What are the correct hypotheses for this test?
The null hypothesis is $H_0: \sigma = 0.006$.
The alternative hypothesis is $H_1: \sigma < 0.006$.

Calculate the value of the test statistic
$\chi^2 = 11.293$ (Round to three decimal places as needed.)

Use technology to determine the P-value for the test statistic
The P-value is 
(Round to three decimal places as needed.)
        
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The piston diameter of a certain hand pump is 0.5 inch. The manager determines that the diameters are normally distributed, with a mean of 0.5 inch and a standard deviation of 0.006 inch. After recalibrating the production machine, the manager randomly selects 22 pistons and determines that the standard deviation is 0.0044 inch. Is there significant evidence for the manager to conclude that the standard deviation has decreased at the α = 0.10 level of significance?

What are the correct hypotheses for this test?
The null hypothesis is H0: σ = 0.006.
The alternative hypothesis is H1: σ < 0.006.

Calculate the value of the test statistic
χ^2 = 11.293 (Round to three decimal places as needed.)

Use technology to determine the P-value for the test statistic
The P-value is 
(Round to three decimal places as needed.)

Added by Daniel M.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Her significant evidence for the manager to conclude that the standard deviation has decreased at the 10% significance level. What are the correct hypotheses for this test? The null hypothesis is H0: σ = σ0. The alternative hypothesis is H1: σ < σ0. Calculate the value of the test statistic as 1.295 (Round to three decimal places as needed). Use technology to determine the P-value for the test statistic. The P-value is [P-value]. (Round to three decimal places as needed).
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Transcript

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00:01 Alright, so in the given question, we have been asked to conduct a hypothesis test.
00:05 So here our null hypothesis is that population average is less than equal to 10.
00:11 And the alternative hypothesis is that population average is greater than 10.
00:15 So here we have a right -tailed hypothesis test.
00:21 Now here the size of the sample is 10.
00:24 The sample mean is 12 and sample standard deviation is 3.
00:28 Right so here the population standard deviation is unknown which means that we are going to use the t test right so first of all we have been asked to determine the decision rule for here so alpha here we assumed is 0 .05 and the degrees of freedom here is 10 minus 1 that is 9 so our critical value corresponds to the value of t alpha for the given degrees of freedom which is t 0 .059 and this is equal to 1 .833 right so the decision rule here is that we will reject a null hypothesis if the test statistic value is greater than 1 .833 now the formula to compute the test statistic value this is equal to sample mean minus population mean divided by standard deviation upon square root of sample size so this is equal to 12 minus 10 upon 3 divided by square root of 10...
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