Question 364:

Asked on September 4, 2008

Tags: Sample Size , Confidence Intervals , Margin of Error

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Answer:

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1. Q364_1_q364.xls

The first thing you should do is get the mean and standard deviation of the sample of pages. You can use Excel, and use the formula =STDEV() for the standard deviation and =AVERAGE() for the mean. You should get a mean of 346.5 and standard deviation of 170.4.

First I will answer c, which will then answer a. We need a sample size which will generate a margin of error of +/-10 sq millimeters. The margin of error is calculated by multiplying the standard error of the mean (SEM) by the critical value of the t or z-distribution. As the sample size increase the values of t and z converge (esp. above 30). For part c I'll use the z-distribution then use the t-distribution on part a since the sample if 20 is relatively small. For a two-sided 99% confidence interval, the z-score is 2.58, which you lookup using the percentile to z-score calculator (enter .01).

We can then just setup and equation and solve for the unknown:

1. The margin of error is the standard error of the mean SEM * the critical z value of  2.58, so working backwards from a margin of 10, the SEM = 10/2.58= 3.8822.
2. The SEM  of 3.88 is made up of the standard deviation divided by the square root of the needed sample size = SD/ SQRT(n). Solving for the unknown sample size n and using the standard deviation we obtained from the sample of 20. We have the equation 170.4 /sqrt(n) = 3.8822
3. Squaring both sides = 29036/n  = 15.07148
4. Isolating n gets us 15.07148n= 29036 or n = 29036/15.07148= 1926.5
5. Rounding up to the nearest whole page gets us 1927.

So a huge sample size of 1927, will generate a  99% +/- 10 mm margin of error.  It should be obvious why this is a problem since the required sample is larger than the entire phone book!

Normality will likely be an issue since there are a couple extreme values (0's) in the data, which will affect the accuracy of the confidence intervals.

Lets now generate a 95% CI around the mean, since we're using the small sample of 20, I will switch and use the t-distribution (which is better for small samples).

1. Calculate the SEM as the SD/SQRT(n) = 170.4/4.47 = 38.1
2. Find the critical value from the t-distribution using the Excel formula =TINV(.05,19), where 19 is the sample size minus 1 and .05 is for a 95% CI. You should get 2.093.
3. Calculate the Margin of Error by multiplying the SEM by the critical t value = 38.1*2.093 = 79.74
4. The end-points of the CI are the mean +/- the margin or 346.5-79.74 and 346.5+79.74 gets a 95% CI of between  266.8 and 426.2

So we can now see that a sample of 20 generates a margin of error of around 80mm for a 95% CI vs. a margin of error of 10 from one that requires a sample of 1929. A reasonable approach to me would be doubling or tripling the sample to 40-60 and using a 95% CI to get a decent idea of the ad space in the phone book.

Note: you might not have covered the t-distribution in your course and might just have covered the z or normal distribution. If that's the case, then the only difference is the critical value found in step 2. It would be 1.959 instead of 2.093. That would make the confidence interval between 271.88 and 421.17. You can see that its pretty close to the t-method and the intervals are a bit narrower.