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Question 451:



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  1. Calculate the standard error of the mean: standard deviation / square root of the sample size = 15/SQRT(35) =2.54
  2. Find the critical value from the t-distribution for 34 degrees of freedom and a probability of .05 (since this is for a 95% confidence interval). You can use MS Excel and type =TINV(.05,34) and should get 2.03.
  3. Calculate the margin of error: which is the standard error of the mean times the t-value =2.54*2.03 = 5.15
  4. Generate the Lower and Upper bounds of the confidence interval by adding and subtracting the margin of error to the mean = 82 +5.15= 87.15 and  82 -5.15  = 76.8.
  5. So your 95% confidence interval around the mean is (76.8, 87.15).

Using the same method above, the only difference for a 99% confidence interval is to substitute the new critical value in step 2 with =TINV(.01,34) = 2.73. Which gives you a new margin of error of 6.9. You should get a 99% CI of between (75.1, 88.9).

You can see that the 99% Confidence Interval is wider since we're asking for a higher level of confidence. As the confidence level increases the width of our interval also increases.

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