Question

In a recent survey, it was found that 133 out of 631 randomly selected Knott's Berry Farm attendees consider Silver Bullet their favorite ride at Knott's. It was also found that 23 out of 121 randomly selected Knott's Berry Farm employees consider Silver Bullet their favorite ride. Test the hypothesis that the percentage of attendees who consider Silver Bullet their favorite ride is different than the percentage of Knott's Berry Farm employees who consider Silver Bullet their favorite ride. Use the a=.05 level of significance.
The P-value is __________, rounded to the nearest ten-thousandth (4 decimal places).

Answer 1

Using the z-distribution, it is found that the p-value is of 0.5975.

At the null hypothesis, it is tested if the proportions are equal, that is, their subtraction is 0, hence:

$H_0: p_1 - p_2 = 0$

At the alternative hypothesis, it is tested if they are different, that is, their subtraction is not 0, hence:

$H_1: p_1 - p_2 \neq 0$

The sample sizes and proportions are given by:

$n_1 = 631, p_1 = \frac{133}{631} = 0.2108$

$n_2 = 121, p_2 = \frac{23}{121} = 0.1901$

The standard errors are given by:

$s_1 = \sqrt{\frac{0.2108(0.7892)}{631}} = 0.0162$

$s_2 = \sqrt{\frac{0.1908(0.8092)}{121}} = 0.0357$

For the distribution of differences, the estimate and the standard error are:

$\overline{p} = p_1 - p_2 = 0.2108 - 0.1901 = 0.0207$

$s = \sqrt{s_1^2 + s_2^2} = \sqrt{0.0162^2 + 0.0357^2} = 0.0392$

The test statistic is given by:

$z = \frac{\overline{p} - p}{s}$

In which $p = 0$ is the value tested at the null hypothesis.

Then:

$z = \frac{\overline{p} - p}{s}$

$z = \frac{0.0207 - 0}{0.0392}$

$z = 0.528$

Using a z-distribution calculator, with a two-tailed test, as we are testing if the mean is different of a value, the p-value is of 0.5975.

A similar problem is given at brainly.com/question/25869410