Question

A manufacturer uses a new production method to produce steel rods. A random sample of 17 steel rods resulted in lengths with a standard deviation of 2.1 cm. At the 0.05 significance level, test the claim that the new production method has lengths with a standard deviation different from 3.5 cm, which was the standard deviation for the old method. Use the p-value method.

Initial Claim:

Null Hypothesis:

Alternative Hypothesis

Test statistic (make sure you state which test statistic that you are using):

P- Value

Initial conclusion (justify your answer – graphs are acceptable):

Final conclusion:

Answer 1

Solution:

Given:

Sample size = n = 17

Sample standard deviation = s = 2.1 cm.

Level of significance =

Population standard deviation = cm.

Part a) Initial Claim:

The new production method has lengths with a standard deviation different from 3.5 cm.

Part b) Null Hypothesis and Alternative Hypothesis:

Since claim is non-directional , the test is two tailed and stating about the standard deviation, we have following null and alternative hypothesis:

Vs

Part c) Test statistic

For testing of hypothesis about population variance or standard deviation of one population, we use Chi-square test statistic of variance or standard deviation and it has following formula:

Part d) P- Value

To get exact P-value, we use excel command:

=CHISQ.DIST.RT( chi-square test statistic , df )

where df = n - 1 = 17 - 1 = 16

Thus

=CHISQ.DIST.RT( 5.76 , 16)

=0.9905

P-value = 0.9905

We can use Chi-square table to get range of P-values:

fall between 5.142 and 5.812

its corresponding right tail area is between 0.990 and 0.995

Thus range of P-value is:

0.9900 < P-value < 0.9950

Part e) Initial conclusion

Since P-value = 0.9905 > Level of significance = , we fail to reject H0.

Part f) Final conclusion:

Since we failed to reject H0, there is not sufficient evidence to support the claim that the new production method has lengths with a standard deviation different from 3.5 cm.