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The piston diameter of a certain hand pump is 0.7 inch. The manager determines that the diameters are normally​ distributed, with a mean of 0.7 inch and a standard deviation of 0.003 inch. After recalibrating the production​ machine, the manager randomly selects 25 pistons and determines that the standard deviation is 0.0022 inch. Is there significant evidence for the manager to conclude that the standard deviation has decreased at the α=0.05 level of​ significance?

Sagot :

Answer:

The answer is "yes".

Step-by-step explanation:

Given:

[tex]n= 25\\\\s = 0.0025\\\\\sigma_{0}= 0.005\\\\\alpha=0.01\\\\[/tex]

Claim:

[tex]\sigma < 0.42[/tex]

Its assertion is the null presumption or the alternative. The hypothesis should include the equal value of the claim. If indeed the claim is indeed the null assumption, the alternative hypothesis is indeed the opposite.    

[tex]H_o: \sigma =0.42\\\\H_1 : \sigma<0.42[/tex]  

Calculating the test statistic:

[tex]X^2=\frac{n-1}{\sigma_{0}^2} s^2=\frac{25-1}{0.005^2} \times 0.0025^2 \approx 6[/tex]

The key value of the left-tails test is shown in the row [tex]df = n -1=25-1 = 24[/tex]  and the column with [tex]1 -\alpha = 0.99[/tex] in an appendix, from the chi-square table:

[tex]X_{1-\alpha}^{2}= 10.856[/tex]

All values are smaller than 10.856 in the reject region.

Unless the test statistic is in the refusal region, this same null hypothesis is rejected:

[tex]6 < 10.856 \to Reject \ H_{0}[/tex]

Its claim that perhaps the recalibration is efficient is demonstrated by sufficient evidence.