Question

A survey was conducted among people who visit swimming pools. 209 people were asked about the preferences in water sanitizing. 112 respondents reported they prefer pools with mechanical methods of water sanitizing, and 97 respondents reported they do not have any preferences. Construct a test for a difference in proportions for this survey to find out if the proportion of people who prefer mechanical methods of water sanitizing truly exceeds the proportion of people who have no sanitizing methods preferences, if it is possible. A :

Answer 1

Answer:

The claim that needs to be tested is that the proportion of people who prefer mechanical methods of water sanitizing truly exceeds the proportion of people who have no sanitizing methods preferences. This can be expressed as the population proportion of people who prefer mechanical methods of water sanitizing is significantly higher than 0.5.

This is a hypothesis test for a  population proportion.

The test will have null and alternative hypothesis:

$H_0: \pi=0.5\\\\H_a:\pi>0.5$

Test statistic z = 0.972

P-value = 0.166

At a significance level of 0.05, there is not enough evidence to support the claim that the proportion of people who prefer mechanical methods of water sanitizing truly exceeds the proportion of people who have no sanitizing methods preferences.

Step-by-step explanation:

We have to construct a hypothesis test for the population proportion. We need to test the claim that the proportion of people who prefer mechanical methods of water sanitizing truly exceeds the proportion of people who have no sanitizing methods preferences.

As the survey is done as a yes/no question, the null hypothesis would state that the amount of people who prefer mechanical methods of water sanitizing is not significantly different from the proportion of people who have no sanitizing methods preferences. That can be expressed as the population proportion equals 0.5, so both proportions are equal.

The alternative hypothesis would state that the proportion of people who prefer mechanical methods of water sanitizing truly exceeds the proportion of people who have no sanitizing methods preferences. This could be expressed as the population proportion significantly bigger than 0.5.

Then, the null and alternative hypothesis are written:

$H_0: \pi=0.5\\\\H_a:\pi>0.5$

The significance level is assumed to be 0.05.

The sample has a size n=209.

The sample proportion is p=0.536.

$p=X/n=112/209=0.536$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.5*0.5}{209}}\\\\\\ \sigma_p=\sqrt{0.001196}=0.035$

Then, we can calculate the z-statistic as:

$z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.536-0.5-0.5/209}{0.035}=\dfrac{0.034}{0.035}=0.972$

This test is a right-tailed test, so the P-value for this test is calculated as:

$\text{P-value}=P(z>0.972)=0.166$

As the P-value (0.166) is greater than the significance level (0.05), the effect is  not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the proportion of people who prefer mechanical methods of water sanitizing truly exceeds the proportion of people who have no sanitizing methods preferences.