7.4.2.3 - Example: 99% CI for Proportion of Women Students

7.4.2.3 - Example: 99% CI for Proportion of Women Students

Scenario: Data were collected from a representative sample of 501 World Campus STAT 200 students. In that sample, 284 students were women and 217 were not women. Construct a 99% confidence interval to estimate the proportion of all World Campus students who are women. 

StatKey was used to construct a sampling distribution using bootstrapping methods:

Left Tail Two - Tail Right Tail Confidence Interval for a Proportion Generate 1 Sample Generate 10 Samples Generate 100 Samples Generate 1000 Samples Edit Data Custom Data Proportion Bootstrap Dotplot of Count 284 501 0.567 Sample Size Proportion Original Sample samples = 5000 mean = 0.567 std. error = 0.022 Reset Plot 125 100 75 200 175 150 50 25 0 0.50 0.52 0.54 0.56 0.58 0.60 0.62 0.64 0.567

Because this distribution is approximately normal, we can approximate the sampling distribution using the z distribution. We will use the standard error, 0.022, from this distribution.

The original sample statistic was \(\widehat p =\frac{284}{501}=0.567\). 

We can find the \(z^*\) multiplier by constructing a z distribution to find the values that separate the middle 99% from the outer 1%:

-2.57583 2.57583 0.005 0.005 Distribution Plot Normal, Mean=0, StDev=1 0.0 0.1 0.2 0.3 0.4 0 X Density

The \(z^*\) multiplier is 2.57583

Recall the general form of a confidence interval: sample statistic \(\pm\) \(z^*\) (standard error) where \(z^*\) is the multiplier. So in this case we have...

\(0.567 \pm 2.57583 (0.022)\)

\(0.567 \pm 0.057\)

\([0.510, 0.624]\)

I am 99% confident that the proportion of all World Campus students who are women is between 0.510 and 0.624

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