10.3: Estimating a Population Proportion • Anytime we wish to know the truth about the percentage of a population, we need to use inference procedures for p the population proportion. • Ex. What proportion of US adults are unemployed? • Ex. What percentage of teenagers have a computer with Internet connection in their bedroom?

•

We have to use what we know about p-hat in order to find probabilities.

•

Center: The mean of the sampling distribution of p-hat is the true proportion p. p(1  p) Spread: The std. dev. of p-hat is if n the population is ten times the sample. Shape: If the sample size is large enough that both np and n(1 - p) are greater than 10, the distribution of p-hat is approximately normal.

• •

• However, in practice, we do not know the value of p. (If we did, we would not need to construct a confidence interval!) – We cannot check if np and n(1 – p) are greater than 10.

• We must replace p with p-hat when we check this condition.

• We also cannot use p to find the standard deviation of p-hat so we will have to replace that too. – When you replace a parameter with an estimate, it is now called the standard error of p-hat.

SE pˆ 

pˆ 1  pˆ  n

•

Conditions for Inference about a Proportion

1. SRS 2. Normality: p-hat ~ N •

Check n(p-hat) and n(1 – p-hat) are greater than 10

3. Independence: Individual observations are independent. When sampling without replacement, check that the population is at least 10 times as large as the sample.

• Our only option is to check that p-hat is approximately normal, so all confidence intervals for a proportion will be z intervals. • Confidence Interval for a Population Proportion: draw an SRS of size n from a large population with unknown proportion p of successes. An approximate level C confidence interval for p is: ˆ (1  p ˆ) p * ˆz p

n

Where z* is the upper (1 – C)/2 critical value of the standard Normal distribution.

• Ex. The Harvard School of Public Health survey found that 2486 of a sample of 10,904 college undergraduates said they had engaged in frequent binge drinking. Construct and interpret a 99% confidence interval for the proportion of college undergraduates who frequently binge drink. Step 1:

Step 2:

Step 3:

Step 4:

• Note: Margin of error in this interval only includes random sampling error. For example, we are forced to assume that everyone told the truth. Also, there is no way to account for nonresponse. Of all the surveys sent out a little over 50% were returned. This is actually a very high number for surveys – but the question remains, what would those other 50% of students have said?

Choosing a Sample Size • The margin of error for a confidence interval of p is:

me  z

*

ˆ 1  p ˆ p n

• Therefore, knowing the confidence level will tell you z*, but what about p-hat? You haven’t done the survey yet to know what p-hat is!

• Option 1: Use a guess, p*, based on a pilot study or on past experience with similar studies. – Ex. If the President’s approval rating was at 38% last month, use that as your guess for p* to find the sample size.

• Option 2: Use p* = 0.5 as the guess. The margin of error, me, is largest when p-hat = 0.5, so that will be a conservative guess. Any other value of p-hat will give us less margin of error than we have planned for.

• To determine the sample size n that will yield a level C confidence interval for a population proportion p with a specified margin of error me, set the following expression for the margin of error to be less than or equal to me and solve for n.

z

*



p 1 p n *

*

  me

• Where p* is the guessed value for the sample proportion.

• Changing p* doesn’t change n too much as long as p* is not too far from 0.5. – Use the conservative guess p* = 0.5 if you expect the true p to be roughly between 0.3 and 0.7. – If you expect p to be closer to 1 or 0, then you should try using a better guess from a pilot study.

• A company has received complaints about its customer service. They intend to hire a consultant to carry out a survey of customers. Before contacting the consultant, the company president wants some ideas of the sample size the she will be required to pay for. One critical question is the degree of satisfaction with the company’s service, measured on a fivepoint scale. The president wants to estimate the proportion p of customers who are satisfied (that is, who choose either “satisfied” or “very satisfied,” the two highest levels on the five-point scale). She decides that she wants the estimate to be within 3% at a 95% confidence level. What sample size will be required?

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A poll on MSNBC contacted 1019 adults and surveyed if they believed that efforts to promote equality for women have gone far enough. An article about the poll said, “Results have a margin of sampling error of plus or minus 3 percentage points. 1. Overall, 54% of the sample (550 of 1019) answered “Yes.” Construct and interpret a 95% confidence interval for the proportion in the adult population who would say “Yes” if asked. Is the report’s claim about the margin of error roughly correct. (Assume the sample is an SRS.)

Step 1:

Step 2:

Step 3:

Step 4:

2. The news article said that 65% of men, but only 43% of women, think that efforts to promote equality have gone far enough. Explain why we do not have enough information to give confidence intervals for men and women separately.

3. Would a 95% confidence interval for women alone have a margin of error less than 0.03, about equal to 0.03, or greater than 0.03? Why?

4. What size sample should they use if they wished to be 99% confident in their results?