Confidence Intervals and Tests of Proportions

Assumptions for inference when using sample proportions:

We will develop a short list of assumptions for making confidence intervals and tests of significance with proportions.

As with all other statistical inference, we will need to have simple random samples when working with proportions.

This insures that we have unbiased samples of our target populations.

(Remember that statistics cannot correct a poor experimental or sampling design. Planning prior to collecting data is the best way to be sure that all assumptions will be met.)

Our first assumption is that we have a simple random sample.

When working with proportions we will use a normal approximation and hence, the Z distribution.

We have requirements for using the normal distribution. We encountered these earlier when working with sampling distributions of sample proportions.

In order to use our formula for standard deviation we need to know that our population is large relative to our sample. (There is a more complicated formula for standard deviation when the sample makes up a large part of the population.)

Rule of Thumb #1: The population is ten times (or more) as large as our sample.

In order to justify the use of the normal approximation we apply another set of rules:

Rule of Thumb #2: np≥10 and n(1-p)≥10.

For significance tests we will use p0 to estimate p.

For confidence intervals we will use ˆ p population proportion.

to estimate p, the true

Problem: Have efforts to promote equality for women gone far enough in the United States? A poll on this issue by the cable network MSNBC contacted 1019 adults. A newspaper article about the poll said, “Results have a margin of sampling error of plus or minus 3 percentage points.”

Overall, 54% of the sample (550 of 1019) answered “Yes.” Find a 95% confidence interval for the proportion in the adult population who would say “Yes” if asked. If the sample is an SRS, is the report’s claim about the margin of error roughly correct?

Step 1: Z interval for proportions

Step 2: Assumptions:

•We have an SRS, given.

•The population is more than 10 times the sample. (The adult US population is well over (10)(1019).

nˆ p =1019⋅.54≈550≥10•

n(1−ˆ p )=1019⋅.46≈469≥10•

Step 3: ˆ p ±z* ˆ p (1−ˆ p )n

= .54±1.960.54⋅.461019

=.54±.0306

€

(.51,.57)or

Step 4: We are 95% confident that the true proportion of US adults who would answer “Yes” to the stated question is between 51% and 57%.

or

In repeated sampling, the confidence interval calculation we have used will give results that capture the true proportion 95% of the time.

The margin of error is about 3%, as stated.

Now we will write up a test of significance testing a proportion.

Problem: At a 5% level of significance, is the proportion of US adults who answer “Yes” to the stated question significantly greater than 50%?

Step 1:

H0: p = .5

Ha: p > .5

The proportion of US adults who answer “Yes” if asked if the US has done enough to promote equality for women is 50%.

The proportion of US adults who answer “Yes” if asked if the US has done enough to promote equality for women is greater than 50%.

Step 2: Assumptions:

• We are given an SRS.

• The population is more than 10 times the sample size.

• np0 =1019⋅.5=509.5≥10

• n(1−p0)=1019⋅ .5=509.5≥10

Step 3: z=ˆ p −p0

p0(1−p0)n

=.54−.5.54⋅.461019

= 2.56

Step 4:

Step 5: P-value = P(z > 2.56) = .00523

Step 6: Reject H0, a value this extreme will occur by chance alone less than 1% of the time.

Step 7: We have strong evidence that the proportion of US adults who answer “Yes” to a question of whether the US has done enough to promote equality for women is greater than 50%.

THE END