Confidence intervals

When deriving a confidence interval for the population mean, we had to be concerned about the sampling distribution of $\overline{X}$ , i.e. how $\overline{X}$ can vary among repeated samples from the same population.

In a similar fashion, a confidence interval for the population proportion can also be found.

Example:

A polling organization use random digit dialing to select a sample of 1089 adult Canadians. Of these, 212 favor the Rhinoceros Party. What can be said about the true level of support among Canadians.

How do we proceed?

• The population parameter to be estimated is p, the proportion of ALL adult Canadians who are in favor of the Rhinoceros Party. The value of p is unknown, and could not be known unless every single Canadian is surveyed. This is obviously too expensive and not feasible to do. [Note this is analogous to sampling for a mean, where $\mu$ was the unknown quantity.

• The sample statistics are that n=1089 (the number of Canadians surveyed). Of the 1089 people surveyed, 212 out of 1089 = .195 = 19.5% were in favor of the R.P. and (1089-212) = 877 out of 1089 = .805 = 80.5% were not in favor of the R.P. These are sample statistics and we use the symbol $\widehat{p}$ to refer to the sample proportion observed. [Refer back to Assignment 6 where you had to survey the proportion of cars in the parking lot - the same symbol was used.]

• This can be entered in JMP as shown below:
  

• Then use Analyze->Distribution-of-Y to analyze the `response' variables. Use the pop-down menu in the frequency chart to display the estimated standard error for each response as shown below:
  

  

• The sample proportions (the $\widehat{p}$ - one for each level) are 19.5% and 80.5% respectively.

• The estimated standard errors are found as noted above. For example: the estimated standard error for the sample proportion of `in favor' is $\sqrt{\frac{.195(1-.195)}{1089}}$ = .0012. This is an estimate of the standard deviation of $\widehat{p}_{in favor}$ over all possible repeated samples drawn from this population.

• Using exactly the same logic as when we found a confidence interval for the mean, a confidence interval for the population proportion is found (approximately) as

  $\widehat{p}$ $\pm$ 2 estimated se of $\widehat{p}$ or

  $\widehat{p} \pm 2 \times \sqrt{\frac{\widehat{p} (1-\widehat{p})}{n}}$

  In the above example, an estimated confidence interval for $p_{in favor}$ is found as 195 $\pm$ 2 (.012) = .195 $\pm$ .024 = (.171 to .219) = (17.1% to 21.9%).

• We interpret this interval as `being 95% confident that the true proportion of Canadians in favor of the R.P is somewhere between 17.1% and 21.9%'.

• Another way of saying this is that `the results are accurate to within 2.4 percentage points, 19 times out of 20' or that `the margin of error is 2.4 percentage points, 19 times out of 20'.

Additional details and Cautions

• Be sure to carefully distinguish between p and $\widehat{p}$ .

• Note that the confidence interval is computed using $\widehat{p}$ , but is a confidence interval for p - the population parameter that is unknown.

• Be careful to distinguish between a percentage point and a percent. In the above we say that the margin of error is 2.4 percentage points, i.e., within $\pm$ .024. If you state that the results are accurate to within 2.4 percent, you really are saying that the margin of error is $\pm$ (.024 of .195) or $\pm$ .00468 which is not correct. This is a common error often made in newspapers or on TV - watch Peter Mansbridge on The National make this mistake!

• The above formula are only valid under simple random sampling. It is not valid for convenience samples or any other type of sampling that is not a simple random sample! Also, it covers uncertainly introduced by sampling only and does not include errors caused by non-response, biased questions, etc. which are typically much larger. These are similar conditions as for confidence intervals for a mean .

• A more `exact' confidence interval can be found using a t-distribution (again use n-1 degrees of freedom). However, the sample sizes for polls are typically several hundred or larger and at these sample sizes, the t-distribution is so close to 2 that it is pointless trying to refine it further.

• Notice that the formula for the confidence interval depends only on n - the total sample size and NOT ON THE POPULATION SIZE. This was explained earlier using the pot-of-soup analogy. Consequently, a random sample of 1089 people from the US is just as accurate as 1089 from Canada, and a random sample of 1089 people from the WORLD is just as accurate as 1089 from Canada! Incredible as it seems, this is a consequence of a RANDOM sample and the central limit theorem. [The CLT, it may change your life!]

• The estimated se and the margin of error (the $\pm$ part of the confidence interval) will change as $\widehat{p}$ varies. For a fixed sample size, it is biggest when $\widehat{p}$ =.50. The Gallup and Angus-Reid and other pollsters usually give a margin of error of 3 percentage points for sample sizes of 1000 people. They are simply using the worse case that could occur when $\widehat{p}$ .

  is .50. In this case the margin of error is almost exactly 3 percentage points.

Additional examples.

1. In recent years, there has been growing concern about the health effects of video display terminals (VDTs).

   These results were based on a University of Michigan study presented at the annual meeting of the American Public Health Association.

   It is known that the miscarriage rate of a pregnancy under `normal' conditions is about 20%.

   In a survey of 697 part-time female employees who used VDTs 20 hours or less per week, there were 145 reported miscarriages.

   1. What is the point estimate of the miscarriage rate?

      Here $\widehat{p}$ =145/697 = .208.

   2. Find an approximate 95% confidence interval for the miscarriage rate. Interpret this interval.

      We find the estimated se of $\widehat{p}$ = $\sqrt{\frac{\widehat{p} (1- \widehat{p})}{n}}$ = $\sqrt{\frac{.208(1-.208)}{697}}$ = .0154.

      An approximate 95% confidence interval is found as: $\widehat{p}$ $\pm$ 2 se of $\widehat{p}$ = .208 $\pm$ 2(.0154) = .208 $\pm$ .0308 = (.177 to .239).

      We are 95% confident that the true miscarriage rate for women who work 20 hours or less per week with VCDTs is in this interval.

   3. Compare the miscarriage rate of works with the rest of the population:

      Because the 95% confidence interval includes the hypothesized value of 0.20, there is no evidence that the miscarriage rate has changed for this group of workers.

2. An ornamental horticulturist has developed a new variety of grass seed for dry climates (less than 10 in or rain per year). Testing on 150 plots revealed that 70% of seed germinated in dry conditions. What does this tell you about the true germination rate?

3. Is there a relationship between weather and violent crime? In the paper `Is there a season for homicide'; (Criminology, 1988), the author classified 1361 homicides by season as follows:
   

   Winter	328
   Spring	334
   Summer	372
   Fall	327