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Computing the Standardize Test Statistic
You are testing the claim that the proportion of men who own cats is significantly proportion of women who own cats. Assume the null hypothesis is that the popul equal.
You sample 100 men, and \( 80 \% \) own cats.
You sample 180 women, and \( 85 \% \) own cats.
Find the test statistic, rounded to two decimal places.
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3. Calculate test statistic: \(Z = \dfrac{(p_{1} - p_{2}) - 0}{\sqrt{\frac{p(1-p)}{n_{1}} + \frac{p(1-p)}{n_{2}}} } = \dfrac{(0.80-0.85) - 0}{\sqrt{\frac{\dfrac{233}{280}(1-\dfrac{233}{280})}{100} + \frac{\dfrac{233}{280}(1-\dfrac{233}{280})}{180}} } =\dfrac{-0.05}{\sqrt{\frac{233(47)}{280(100)} + \frac{233(47)}{280(180)}} } = -1.60\)
Step 1 ：1. \(p_{1} = 0.80\), \(n_{1} = 100\), \(p_{2} = 0.85\), \(n_{2} = 180\)
Step 2 ：2. Compute pooled proportion: \(p = \dfrac{p_{1}n_{1} + p_{2}n_{2}}{n_{1} + n_{2}} = \dfrac{0.80(100) + 0.85(180)}{100+180} = \dfrac{80+153}{280} = \dfrac{233}{280}\)
Step 3 ：3. Calculate test statistic: \(Z = \dfrac{(p_{1} - p_{2}) - 0}{\sqrt{\frac{p(1-p)}{n_{1}} + \frac{p(1-p)}{n_{2}}} } = \dfrac{(0.80-0.85) - 0}{\sqrt{\frac{\dfrac{233}{280}(1-\dfrac{233}{280})}{100} + \frac{\dfrac{233}{280}(1-\dfrac{233}{280})}{180}} } =\dfrac{-0.05}{\sqrt{\frac{233(47)}{280(100)} + \frac{233(47)}{280(180)}} } = -1.60\)
