# In a recent survey of gun control laws, a random sample of 1000 women showed that 65% were in...

## Question:

In a recent survey of gun control laws, a random sample of {eq}1000 {/eq} women showed that {eq}65\% {/eq} were in favour of stricter gun control laws. In a random sample of {eq}1000 {/eq} men, {eq}60\% {/eq} favoured stricter gun control laws. Construct a {eq}95\% {/eq} confidence interval for {eq}\displaystyle p_1 - p_2 {/eq}.

## Confidence Interval

The confidence interval for the difference in proportions can be used to conduct a hypothesis test. If the null hypothesis value is contained in the boundaries of the confidence interval, the null hypothesis is not rejected.

The answer is the 95% confidence interval of the difference in the proportions is:

{eq}\pi_{diff} = 0.05 \pm 0.0424\ \ \text{or}\ [0.0076, 0.0924] {/eq}

To create the confidence interval, we find the Z score for {eq}95\% {/eq} probability. The Z look up table found in a textbook or on the internet is set up to give the one tail of the probability distribution. The confidence interval is a two-tail probability. Look up {eq}\dfrac{(1 + 0.95)}{ 2} = 0.975{/eq} to get the Z score for the {eq}95\% {/eq}confidence interval, and find Z is {eq}1.96 {/eq}.

The equation for the confidence interval estimate for the difference between two proportions is:

{eq}\pi_{diff} = (p_{1} - p_{2}) \pm Z \sqrt{\dfrac{p_{1}(1 - p_{1})}{n_{1}} + \dfrac{p_{2}(1- p_{2})}{n_{2}}}{/eq}

Where:

• {eq}\pi_{diff}{/eq} is the differences in the population proportions
• {eq}p_{1}{/eq} is the proportion of females = 650 / 1,000 = 0.65
• {eq}p_{2}{/eq} = the proportion of males = 600 / 1,000 = 0.60
• {eq}Z{/eq} is the Z score = 1.96
• {eq}n_{1}{/eq} is the sample size of females = 1,000
• {eq}n_{2}{/eq} is the sample size of males = 1,000

Substituting values:

{eq}\begin{align} \pi_{diff} &= (0.65 - .60) \pm 1.96 \sqrt{\dfrac{0.65 * 0.35}{1,000} + \dfrac{0.60 * 0.40}{1,000}}\\ \pi_{diff} &= 0.05 \pm 0.0424\ \ \text{or}\ [0.0076, 0.0924] \end{align}{/eq}