Question

in the recent census, three percent of the u.s. population reported being of two or more races. however, the percentage varies tremendously from state to state. suppose that two random surveys are conducted. in the first random survey, out of 1,000 north dakotans, only nine people reported being of two or more races. in the second random survey, out of 500 nevadans, 17 people reported being of two or more races. conduct a hypothesis test to determine if the population percents are the same for the two states or if the percent for nevada is statistically higher than for north dakota.
which distribution would you use for this hypothesis test?
a. normal
b. students t

Explanation:

Step1: Check sample size conditions

For hypothesis testing of two - proportion z - test (which is based on the normal distribution), we need to check \(n_1p_1\geq5\), \(n_1(1 - p_1)\geq5\), \(n_2p_2\geq5\), and \(n_2(1 - p_2)\geq5\).
Let \(n_1 = 1000\), \(x_1=9\), so \(\hat{p}_1=\frac{9}{1000}=0.009\). Then \(n_1\hat{p}_1=9\geq5\) and \(n_1(1 - \hat{p}_1)=1000\times(1 - 0.009)=991\geq5\).
Let \(n_2 = 500\), \(x_2 = 17\), so \(\hat{p}_2=\frac{17}{500}=0.034\). Then \(n_2\hat{p}_2=17\geq5\) and \(n_2(1 - \hat{p}_2)=500\times(1 - 0.034)=483\geq5\).

Step2: Recall the distribution for two - proportion test

The formula for the test statistic in a two - proportion hypothesis test is \(z=\frac{(\hat{p}_1-\hat{p}_2)- (p_1 - p_2)}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1}+\frac{1}{n_2})}}\), where \(\hat{p}=\frac{x_1 + x_2}{n_1 + n_2}\). Since the test statistic \(z\) follows a standard normal distribution (when the sample size conditions are met), we use the normal distribution for this hypothesis test.

Answer:

A. normal