9.1.2.1.2 – Example: Same Sex Marriage

9.1.2.1.2 – Example: Same Sex Marriage

Example: Same Sex Marriage

A survey was given to a random sample of college students. They were asked whether they think same sex marriage should be legal. We're going to compare the students who identified as women and men in terms of whether or not they responded "yes" to this question. Of the 251 women in the sample, 185 said "yes." Of the 199 men in the sample, 107 said "yes."

1. Check any necessary assumptions and write null and alternative hypotheses.

For women, there were 185 who said "yes" and 66 who said "no." For men, there were 107 who said "yes" and 92 who said "no." There are at least 10 successes and failures in each group so the normal approximation method can be used.

  • \(\widehat{p}_{w}=\dfrac{185}{251}\)
  • \(\widehat{p}_{m}=\dfrac{107}{199}\)

This is a two-tailed test because we are looking for a difference between women and men, we were not given a specific direction. 

  • \( H_{0} : p_{w}- p_{m}=0 \)
  • \( H_{a} : p_{w}- p_{m}\neq 0 \)
2. Calculate an appropriate test statistic.

\(\widehat{p}=\dfrac{185+107}{251+199}=\dfrac{292}{450}=0.6489\)

\(SE_0=\sqrt{\frac{292}{450}\left ( 1-\frac{292}{450} \right )\left ( \frac{1}{251}+\frac{1}{199} \right )}=0.0453\)

\(z=\dfrac{\frac{185}{251}-\frac{107}{199}}{0.0453}=4.400\)

Our test statistic is \(z=4.400\)

3. Determine a p value associated with the test statistic.

Distribution Plot of Density vs X - Normal, Mean=0, StDev=1

\(P(z>4.400)=0.0000054\), this is a two-tailed test, so this value must be multiplied by two: \(0.0000054\times 2= 0.0000108\)

\(p<0.0001\)

4. Decide between the null and alternative hypotheses.

\(p\leq0.05\), therefore we reject the null hypothesis.

5. State a "real world" conclusion.

There is evidence, in this population of students, that there is a difference between the proportion of women and men who think that same sex marriage should be legal.


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