Question

A precision instrument is guaranteed to be accurate to within 2 units. A sample of four instrument readings on the same object yielded the measurements 352, 350, 350, and 353. Give the attained significance level for testing the null hypothesis σ = 0.7 versus the alternative hypothesis σ > 0.7. (Round your answer to six decimal places.)

Answer 1

`\chi^2 =(4-1)/(0.7^2) 1.5^2 =13.776`

`p_v =P(\chi^2 >13.776)=0.0032`

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

"=1-CHISQ.DIST(13.776,3,TRUE)"

If we compare the p value and the significance level assumed we see that
`p_v <\alpha` so on this case we have enough evidence in order reject the null hypothesis. And we can conclude that the true deviation i significantly higher than 0.7

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=4` represent the sample size

We can calculate the sample deviation with this formula:

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

`\alpha` represent the confidence level

`s =1.5` represent the sample variance obtained

`\sigma^2 =0.7` represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population deviation is higher than 0.7, so the system of hypothesis would be:

Null Hypothesis:
`\sigma \leq 0.7`

Alternative hypothesis:
`\sigma >0.7`

Calculate the statistic

For this test we can use the following statistic:

`\chi^2 =(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 =(4-1)/(0.7^2) 1.5^2 =13.776`

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 3. And since is a right tailed test the p value would be given by:

`p_v =P(\chi^2 >13.776)=0.0032`

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

"=1-CHISQ.DIST(13.776,3,TRUE)"

Conclusion

If we compare the p value and the significance level assumed we see that
`p_v <\alpha` so on this case we have enough evidence in order reject the null hypothesis. And we can conclude that the true deviation i significantly higher than 0.7