A newsgroup is interested in constructing a 90% confidence interval for the proportion of all Americans who are in favor of a new Green initiative.

Question

A newsgroup is interested in constructing a 90% confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 569 randomly selected Americans surveyed, 429 were in favor of the initiative. Round answers to 4 decimal places where possible.

a. With 90% confidence the proportion of all Americans who favor the new Green initiative is between __ and __ .

b. If many groups of 569 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. About __ percent of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about __ percent will not contain the true population proportion.

Answer 1

We know that we have a sample of 569 randomly selected people with 429 saying "in favor". Since there are 569 randomly selected people in the sample, the sample size is considered large enough to approximate with a normal distribution (by the Central Limit Theorem). Since we are dealing with proportions, we check for normality which are n(p_s) >= 10 and n(1-p_s) >= 10 where p_s = sample proportion.

For our case, p_s = 429/569 ~ 0.753, so 569(429/569) = 429 >= 10 and 569((569-429)/569) = 569(140/569) = 140 >= 10

Normal approximation for this sample is satisfied.

To construct a 90% confidence interval, we use this formula:

p_s +/- z*sqrt((p_s)(1-p_s)/n)

Since we are using a 90% confidence interval, z* = 1.645. Substituting in, we get that the confidence interval is

0.753 +/- 1.645 * sqrt(0.753 * (1-0.753)/569) = 0.753 +/- 1.645*0.018 = (0.753 - 1.645 * 0.018, 0.753 + 1.645 * 0.018) = (0.753 - 0.0296, 0.753 + 0.0296) = (0.7234, 0.7826)

Regarding intuition behind confidence interval, think about it this way (in the context of our problem):

You want to understand the true proportion of your population that supports the new initiative. However, in practice, it is not easy to go out and survey every single person in the population about their thoughts on the new initiative. Instead, we are using a large enough sample to approximate what we think the true population proportion is. Generally, larger sample sizes tend to minimize the error, which makes your confidence interval narrower.

From the sample in this problem, you have collected one proportion based off of that sample. What if we repeat this experiment a bunch of times. If we plot all of the proportions for each sample in a distribution, you start to get a feel for what the true proportion is. Also, this distribution would be centered around the TRUE population proportion. In our case, if we kept sampling those who support the initiative and plot out all confidence intervals, 90% of such intervals will capture the true proportion of individuals that support the initiative.

Hope this helps! Confidence intervals are a very useful concept and sometimes, it is easily misunderstood