Question

below is a table representing data, measuring the percentage of various groups owning a home

*note: do not round during your calculations of the estimated standard error, because your end result will be way off. once you calculated it round the estimated standard error to two decimals. then use that to plug it into the formula. it is general practice to round the z - score also to two decimals.

test whether there is a significant difference in the proportion of homeowners between 1st and 2nd generation hispanic americans. set alpha at.05

the obtained z test is type your answer. the probability of obtaining this z - test statistic is type your answer. this is type your answer than our alpha level. therefore, we type your answer the null hypothesis and conclude that there type your answer a significant difference in the proportion of homeowners between 1st generation and 2nd generation hispanic americans.

Explanation:

Step1: Calculate the pooled proportion

The formula for the pooled proportion \(p=\frac{n_1p_1 + n_2p_2}{n_1 + n_2}\).
Here, \(n_1 = 899\), \(p_1=0.43\), \(n_2 = 351\), \(p_2 = 0.5\)
\(p=\frac{899\times0.43+351\times0.5}{899 + 351}=\frac{386.57+175.5}{1250}=\frac{562.07}{1250}=0.449656\)

Step2: Calculate the standard error

The formula for the standard error \(SE=\sqrt{p(1 - p)(\frac{1}{n_1}+\frac{1}{n_2})}\)
Substitute \(p = 0.449656\), \(n_1 = 899\), \(n_2 = 351\)
\(SE=\sqrt{0.449656\times(1 - 0.449656)\times(\frac{1}{899}+\frac{1}{351})}\)
\(=\sqrt{0.449656\times0.550344\times(\frac{351 + 899}{899\times351})}\)
\(=\sqrt{0.2474\times\frac{1250}{315549}}\)
\(=\sqrt{\frac{309.25}{315549}}\approx\sqrt{0.00098}\approx0.0313\) (rounded to two decimals \(SE = 0.03\))

Step3: Calculate the z - score

The formula for the z - score \(z=\frac{p_1 - p_2}{SE}\)
Substitute \(p_1 = 0.43\), \(p_2 = 0.5\), \(SE=0.03\)
\(z=\frac{0.43 - 0.5}{0.03}=\frac{- 0.07}{0.03}\approx - 2.33\)

Step4: Find the probability

For \(z=-2.33\), using the standard normal table (or a calculator with a normal distribution function), the probability \(P(Z\lt - 2.33)=0.0099\) and \(P(Z\gt2.33) = 0.0099\). Since it is a two - tailed test, the probability \(P = 2\times0.0099=0.0198\)

Answer:

The obtained \(Z\) test is \(-2.33\). The probability of obtaining this \(Z\) - test statistic is \(0.02\). This is less than our alpha level. Therefore, we reject the null hypothesis and conclude that there is a significant difference in the proportion of homeowners between 1st generation and 2nd generation Hispanic Americans.