Question

in the recent census, three percent of the u.s. population reported being of two or more races. however, the percent 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.
find the p - value.
a. 0.002
b. 0.0002
c. 0.02

Explanation:

Step1: Calculate sample proportions

Let \(n_1 = 1000\), \(x_1=9\), \(n_2 = 500\), \(x_2 = 17\)
The sample proportion for North Dakota \(p_1=\frac{x_1}{n_1}=\frac{9}{1000}=0.009\)
The sample proportion for Nevada \(p_2=\frac{x_2}{n_2}=\frac{17}{500}=0.034\)
The pooled proportion \(p=\frac{x_1 + x_2}{n_1 + n_2}=\frac{9+17}{1000 + 500}=\frac{26}{1500}\approx0.0173\)

Step2: Calculate the test statistic

The formula for the test statistic \(z\) in a two - proportion z - test is \(z=\frac{p_2 - p_1}{\sqrt{p(1 - p)(\frac{1}{n_1}+\frac{1}{n_2})}}\)
Substitute the values:
\(\frac{0.034 - 0.009}{\sqrt{0.0173(1 - 0.0173)(\frac{1}{1000}+\frac{1}{500})}}\)
\(=\frac{0.025}{\sqrt{0.0173\times0.9827\times(\frac{1 + 2}{1000})}}\)
\(=\frac{0.025}{\sqrt{0.0173\times0.9827\times0.003}}\)
\(=\frac{0.025}{\sqrt{0.000051}}\)
\(z\approx3.5\)

Step3: Calculate the p - value

Since this is a one - tailed test (testing if \(p_2>p_1\)), we find \(P(Z>3.5)\)
Using the standard normal distribution table or a calculator, \(P(Z > 3.5)=1 - P(Z\leq3.5)\)
From the standard normal table, \(P(Z\leq3.5)\approx0.9998\)
So \(P(Z>3.5)=1 - 0.9998=0.0002\)

Answer:

B. 0.0002