7.4.1.4 - Example: Proportion of Women Students

Research question: Are more than 50% of all World Campus STAT 200 students women?

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. 

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

Randomization Dotplot of Null hypothesis: p = Proportion 0.5 Left Tail Two - Tail Right Tail 200 150 0.44 0.46 0.48 0.50 null = 0.5 0.52 0.54 0.56 0.58 100 50 0

Because this randomization distribution is approximately normal, we can find the p value by computing a standardized test statistic and using the z distribution.

Step 1: Check assumptions and write hypotheses

The assumption here is that the sampling distribution is approximately normal. From the given StatKey output, the randomization distribution is approximately normal. 

\(H_0\colon p=0.50\)
\(H_a\colon p>0.50\)

2. Calculate the test statistic

\(test\;statistic=\dfrac{sample\;statistic-hypothesized\;parameter}{standard\;error}\)

The sample statistic is \(\widehat p = 284/501 = 0.567\).

The hypothesized parameter is the value from the hypotheses: \(p_0=0.50\).

The standard error on the randomization distribution above is 0.022.

\(test\;statistic=\dfrac{0.567-0.50}{0.022}=3.045\)

3. Determine the p value

We can find the p value by constructing a standard normal distribution and finding the area under the curve that is more extreme than our observed test statistic of 3.045, in the direction of the alternative hypothesis. In other words, \(P(z>3.045)\):

3.045 0.0011634 Distribution Plot Normal, Mean=0, StDev=1 0.0 0.1 0.2 0.3 0.4 0 X Density

Our p value is 0.0011634

4. Make a decision

Our p value is less than or equal to the standard 0.05 alpha level, therefore we reject the null hypothesis.

5. State a "real world" conclusion

There is evidence that the proportion of all World Campus STAT 200 students who are women is greater than 0.50.