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Question 486:



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  1. The null hypothesis is the resolution in standard deviations is equal to or greater than 2.
  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.
  3. 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.
  4. 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
  5. We now find the upper bounds of the sample standard deviation by:
    1. Multiply the sample variance times the degrees of freedom : 2.13*29 = 61.77
    2. Divide this by the Chi-Square critical value : 61.77/14.26 = 4.33
    3. 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.

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