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.

a. determine the sample proportions.
determine the sample proportion $\hat{p}_1$.
$\hat{p}_1 = 0.111$ (type an integer or a decimal. round to three decimal places as needed.)
determine the sample proportion $\hat{p}_2$.
$\hat{p}_2 = 0.278$ (type an integer or a decimal. round to three decimal places as needed.)
determine the pooled sample proportion $\hat{p}_p$.
$\hat{p}_p=\square$ (type an integer or a decimal. round to three decimal places as needed.)

Explanation:

Step1: Recall the formula for pooled sample proportion

The formula for the pooled sample proportion \(\hat{p}_p=\frac{x_1 + x_2}{n_1 + n_2}\), where \(x_1\) and \(x_2\) are the number of successes in the two samples, and \(n_1\) and \(n_2\) are the sample sizes.

Step2: Substitute the given values into the formula

Given \(x_1 = 10\), \(n_1=90\), \(x_2 = 25\), \(n_2 = 90\). Then \(\hat{p}_p=\frac{10 + 25}{90+90}\).

Step3: Calculate the value

First, calculate the numerator \(10 + 25=35\), and the denominator \(90 + 90 = 180\). So \(\hat{p}_p=\frac{35}{180}\approx0.194\)

Answer:

\(\hat{p}_p = 0.194\)