## Question 482:

1## Answer:

No answer provided yet.- The null hypothesis is the resolution in standard deviations is equal to or greater than 2.
- To test the hypothesis, we can't use a 1-sample t-test since we want to know if the standard deviation is less than 30. The t-test tests if the mean is lower than a criteria. I'm not aware of a test for a standard deviation against a criteria, but we can construct a confidence interval around the sample standard deviation.
- The sample standard deviation has a Chi-Square distribution with n-1 degrees of freedom, which would give us 29 degrees of freedom here. We need to find the critical value from the Chi-Square distribution for alpha = .01 (one-sided) for degrees of freedom of 29. You can use excel and type =CHIINV(0.99,29) or look the value up in a table. You should get the critical value of 14.26.
- We'll now use the sample variance to compute the confidence interval, which is just the square of the standard deviation, which is 1.46*1.46 = 2.13
- We now find the upper bounds of the sample standard deviation by:
- Multiply the sample variance times the degrees of freedom : 2.13*29 = 61.77
- Divide this by the Chi-Square critical value : 61.77/14.26 = 4.33
- Finally, take the square root of 4.33 to arrive at the upper bounds of the interval =SQRT(4.33) = 2.08

So its close, but using this technique at the 99% level of confidence we cannot reject the null hypothesis since the upper end of the interval exceeds 2.