Question

A manufacturing process drills holes in sheet metal that are supposed to be .5000 cm in diameter. Before and after a new drill press is installed, the hole diameter is carefully measured for 12 randomly chosen parts. At a = .05, can we conclude that the new drill has a diffrent variance?New drill.5005 .5010 .5024 .4988 .4997 .4995.5014 .4995 .4988 .4992 .5042 .4967Old drill.5052 .5053 .4947 .4907 .5031 .4923.5040 .5035 .5061 .4956 .5035 .4962A, What test should I run?B. State the null and AltrenativeC. What is the significance level?D.Find the crital value?E. Compute the value of the statisticF. What is the decision regading the null hypothesis? Intrepret the results

Answer 1

a) F test is a statistical test that uses a F Statistic to compare two population variances, with the sample deviations s1 and s2. The F statistic is always positive number since the variance it's always higher than 0. The statistic is given by:

`F=(s^2_2)/(s^2_1)`

b) H0:
`\sigma^2_1 = \sigma^2_2`

H1:
`\sigma^2_1 \\eq \sigma^2_2`

c)
`\alpha=0.05` represent the significance level provided

Confidence =0.95 or 95%

d) We need to calculate the degrees of freedom first. For the numerator we have
`n_2 -1 =12-1=11` and for the denominator we have

We can find the critical value on the F table or with the following excel code: "=F.INV(1-0.025,11,11)"

And we got
`F_(critc)=3.474`

e)
`F=(s^2_2)/(s^2_1)=(0.0056^2)/(0.0019^2)=8.68`

f) Since our calculated value 8.68 is higher than the critical value of 3.474 we have enough evidence to reject the null hypothesis and at 5 % of significance we can conclude that the two variances are different.

Explanation:

Data given and notation

We can calculate the mean and deviation with the following formulas:

`\bar X =(\sum_(i=1^n X_i))/(n)`

`s= (\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)`

`n_1 = 12` represent the sampe size for the new drill

`n_2 =12` represent the sample size for the old drill

`\bar X_1 =0.500` represent the sample mean for the new drill

`\bar X_2 =0.500` represent the sample mean for the old drill

`s_1 = 0.0019` represent the sample deviation for the new drill

`s^2_1 = 3.76x10^(-6)` represent the sample variance for the new drill

represent the sample deviation for the old drill

`s^2_2 = 3.18x10^(-5)` represent the sample variance for the old drill

Part a

F test is a statistical test that uses a F Statistic to compare two population variances, with the sample deviations s1 and s2. The F statistic is always positive number since the variance it's always higher than 0. The statistic is given by:

`F=(s^2_2)/(s^2_1)`

Solution to the problem

Part B: System of hypothesis

We want to test if the variation for oil stocks it's higher than the variation for utility stocks, so the system of hypothesis are:

H0:
`\sigma^2_1 = \sigma^2_2`

H1:
`\sigma^2_1 \\eq \sigma^2_2`

Part c: Significance level

`\alpha=0.05` represent the significance level provided

Confidence =0.95 or 95%

Part D: Critical value

We need to calculate the degrees of freedom first. For the numerator we have
`n_2 -1 =12-1=11` and for the denominator we have

We can find the critical value on the F table or with the following excel code: "=F.INV(1-0.025,11,11)"

And we got
`F_(critc)=3.474`

Part E :Calculate the statistic

Now we can calculate the statistic like this:

`F=(s^2_2)/(s^2_1)=(0.0056^2)/(0.0019^2)=8.68`

Part F: Decision

Since our calculated value 8.68 is higher than the critical value of 3.474 we have enough evidence to reject the null hypothesis and at 5 % of significance we can conclude that the two variances are different.