Question

the numbers of successes and the sample sizes for independent simple random samples from two populations are provided for a left - tailed test and an 80% confidence interval. complete parts (a) through (d).
( x_1 = 10, n_1 = 90, x_2 = 25, n_2 = 90, alpha = 0.10 )
click here to view a table of areas under the standard normal curve for negative values of z.
click here to view a table of areas under the standard normal curve for positive values of z.
( hat { p } _ { p } = 0.194 ) (type an integer or a decimal. round to three decimal places as needed.)
b. decide whether using the two - proportions z - procedures is appropriate.
check that the assumptions are satisfied. select all that apply.
a. the assumptions are satisfied, so using the procedures is appropriate.
b. since ( n _ { 2 } - x _ { 2 } ) is less than 5, using the procedures is not appropriate.
c. since ( x _ { 2 } ) is less than 5, using the procedures is not appropriate.
d. since ( x _ { 1 } ) is less than 5, using the procedures is not appropriate.
e. since ( n _ { 1 } - x _ { 1 } ) is less than 5, using the procedures is not appropriate.
c. if appropriate, use the two - proportions z - test to conduct the required hypothesis test.
what are the hypotheses for this test?
a. ( h _ { 0 } : p _ { 1 } = p _ { 2 }, h _ { a } : p _ { 1 } > p _ { 2 } )
b. ( h _ { 0 } : p _ { 1 } < p _ { 2 }, h _ { a } : p _ { 1 } = p _ { 2 } )
c. ( h _ { 0 } : p _ { 1 } = p _ { 2 }, h _ { a } : p _ { 1 } < p _ { 2 } )
d. ( h _ { 0 } : p _ { 1 } = p _ { 2 }, h _ { a } : p _ { 1 }
eq p _ { 2 } )
e. ( h _ { 0 } : p _ { 1 } > p _ { 2 }, h _ { a } : p _ { 1 } = p _ { 2 } )
f. ( h _ { 0 } : p _ { 1 }
eq p _ { 2 }, h _ { a } : p _ { 1 } = p _ { 2 } )
g. using the two - proportions z - procedures is not appropriate.

Explanation:

Step1: Check the assumptions for two - proportions z - procedures

For two - proportions z - procedures, we need \(n_1\hat{p}_1\geq5\), \(n_1(1 - \hat{p}_1)\geq5\), \(n_2\hat{p}_2\geq5\), \(n_2(1 - \hat{p}_2)\geq5\)

First, calculate \(\hat{p}_1=\frac{x_1}{n_1}=\frac{10}{90}\approx0.111\), \(\hat{p}_2=\frac{x_2}{n_2}=\frac{25}{90}\approx0.278\)

\(n_1\hat{p}_1 = 10\geq5\), \(n_1(1 - \hat{p}_1)=90 - 10 = 80\geq5\)

\(n_2\hat{p}_2=25\geq5\), \(n_2(1 - \hat{p}_2)=90 - 25 = 65\geq5\)

Since all the values \(n_1\hat{p}_1\), \(n_1(1 - \hat{p}_1)\), \(n_2\hat{p}_2\), \(n_2(1 - \hat{p}_2)\) are greater than or equal to 5, the assumptions for two - proportions z - procedures are satisfied.

Step2: Determine the hypotheses for the left - tailed test

For a left - tailed test comparing two proportions \(p_1\) and \(p_2\), the null hypothesis \(H_0\) is that the two proportions are equal (\(p_1 = p_2\)) and the alternative hypothesis \(H_a\) is that \(p_1

The formula for the pooled proportion \(\hat{p}=\frac{x_1 + x_2}{n_1 + n_2}=\frac{10+25}{90 + 90}=\frac{35}{180}\approx0.194\)

The test statistic \(z=\frac{\hat{p}_1-\hat{p}_2}{\sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1}+\frac{1}{n_2})}}\)

Substitute \(\hat{p}_1 = 0.111\), \(\hat{p}_2=0.278\), \(\hat{p}=0.194\), \(n_1 = 90\), \(n_2 = 90\)

\(z=\frac{0.111 - 0.278}{\sqrt{0.194\times(1 - 0.194)\times(\frac{1}{90}+\frac{1}{90})}}\)

\(=\frac{- 0.167}{\sqrt{0.194\times0.806\times\frac{2}{90}}}\)

\(=\frac{-0.167}{\sqrt{\frac{0.194\times0.806\times2}{90}}}\)

\(=\frac{-0.167}{\sqrt{\frac{0.313}{90}}}\)

\(=\frac{-0.167}{\sqrt{0.00348}}\)

\(=\frac{-0.167}{0.059}\approx - 2.83\)

The critical value for a left - tailed test with \(\alpha = 0.10\) is \(z_{0.10}=-1.28\)

Since \(z=-2.83<-1.28\), we reject the null hypothesis.

Answer:

A. The assumptions are satisfied, so using the procedures is appropriate.

C. \(H_0:p_1 = p_2,H_a:p_1