Applied Research Methods

Spring 2003

Answers to 03/03/03 class exercise

 

3/3/3:  Although we did not have the opportunity to go through this exercise in class, the answers to the first three questions should make sense to you.

 

  1. There are two samples of 100 households each that were designed to represent all the households in Mytown.  The averages generated by these samples are estimates of the household incomes of all the households of the town (i.e., the population).  So, it is not meaningful to compare just the averages without calculating some kind of “margin of error” around them.  In this case we compute the boundaries (limits) of the 95% confidence interval for each mean.  If either mean is contained in the 95% CI of the other, then they cannot be assumed to be different beyond sampling error.

 

Now:

37500 + 1.96 [2000/sqrt (100)] = 37500 + 1.96 (200) = 37500 + 392

           

            95% Confidence Interval = 37108 à 37892

 

Ten years ago:

41000 + 1.96 [2500/sqrt (100)] = 41000 + 1.96 (250) = 41000 + 490

 

            95% Confidence Interval = 40510 à 41490

 

Neither mean is contained in the 95% CI of the other so, we conclude that the average income has changed (decreased) over the past 10 years.

 

  1. Compute the 95% confidence interval for the proportion of votes in favor (P = 58%) found in the sample of 100 voters.  Assuming that more than 50% of the vote is needed to pass the override, then the lower limit of the confidence interval must be greater than 50 for the committee to be “95% confident” that it will pass.

 

.58 + 1.96 [sqrt ((.58)(.42))/100]      = .58 + 1.96 [sqrt (.0024)]

= .58 + 1.96 [.0494]

= .58 + 0.097

 

            95% Confidence Interval = 48.3% à 67.7%

 

This CI indicates tells you to assume that the true proportion of the population that will vote yes, based on your sample, may include values that are at or below 50%; i.e., you cannot assume with confidence that the true proportion will be greater than the 50% needed to pass the override.

 

Because the CI includes 50%, the committee concludes that their sample cannot predict a win beyond the margin of error. 

 

  1. Where do the students with disabilities in Hiddentown fall relative to the state averages for all students in special education?  Note that this is not a sample of Hiddentown students, but the average for the whole population, i.e., all the students in special education in the town.  SKIP, as an advocate for students with disabilities, wants to note how these students do over time in comparison with the trend in the state as a whole.

 

1999 state mean = 280 (SD = 25).  Hiddentown mean = 260.  The difference of –20 from state mean is 20/25 of a standard deviation, or 0.80 SD’s below the mean (Z = -0.80).

 

2001 state mean = 290 (SD = 15).  Hiddentown mean = 270.  The difference of –20 from state mean is 20/15 of a standard deviation, or 1.34 SD’s below the mean (Z = -1.34). 

 

So, you conclude that the apparent 10 point gain in score for the students of Hiddentown is misleading when compared to overall trends across the state and that these students are, in fact, falling further behind.

 

*****

 

  1. This is a tough one because there are so many decisions to make.  There was also a typo on one version of the hand out.  The faculty size (i.e., the population) was supposed to be 5000 (rather than 500).  This doesn’t affect your calculations, but it does give meaning to the question and answer.

 

You first have to decide how much “of a chance” you are willing to take before risking losing money on non-refundable tickets.  Notice that if you choose an error margin of 0, that there is no sample that can deliver (you would have to survey the entire population to measure the mean directly rather than estimating it from a sample of any size). 

 

Once you decide what margin of error (i.e., point spread) you are willing to tolerate, then you compute 95% confidence intervals that have lower limits no lower than the 60% needed to pass the strike vote.  In other words, if you are willing to take a 5% margin of error, then you would solve for the sample size needed to have confidence in a sample in which the proportion of pro-strike votes was 65% (i.e., 60%à 70% 95% CI). 

 

On the other hand, this large margin of error would leave you wondering whether the faculty might strike after all given if your sample proportion wound up to be, say, 56% (95% CI = 51% à 61%).  A week in Bermuda sounds awfully tempting and so you want a small margin of error so you don’t miss out and so you don’t waste your money.  So, let’s pick an error margin of 2% and solve for n.

 

n = (1.96) 2 (.62)(.38) / (.02) 2 = 3.84 (.62) (.38) (.0004) = 2261.76

 

Therefore, our enterprising graduate students would have to survey 2262 faculty members.  In order to be “95% confident” that they should buy their tickets within a 2 point margin of error, their sample of 2262 faculty would have to yield a 62% endorsement of the strike vote. 

 

For comparison, consider what the 95% CI would be if they found 62% of a sample of 100 faculty to be in favor of a strike:  .62 + .095 = 52.9% à 71.5%.  They wouldn’t buy the tickets.