Mathematics, 06.08.2021 16:50 nyny2129

A study was performed to determine the percentage of people who wear life vests while out on the water. A researcher believed that the percentage was different for those who rode jet skis compared to those who were in boats. Out of 400 randomly selected people who rode a jet ski, 86.5% wore life vests. Out of 250 randomly selected boaters, 92.8% wore life vests. Using a 0.10 level of significance, test the claim that the proportion of people who wear life vests while riding a jet ski is not the same as the proportion of people who wear life vests while riding in a boat. Let jet skiers be Population 1 and let boaters be Population 2. Step 2 of 3:
Step 1 of 3:
State the null and alternative hypotheses for the test. Fill in the blank below.
H0Ha: p1=p2: p1⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯p2H0: p1=p2Ha: p1_p2
Step 3 of 3:
Draw a conclusion and interpret the decision.
Compute the value of the test statistic. Round your answer to two decimal places.

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From the test the person wants, and the sample data, we build the test hypothesis, find the test statistic, and use this to reach a conclusion.

This is a two-sample test, thus, it is needed to understand the central limit theorem and subtraction of normal variables.

Doing this:

The null hypothesis is $H_0: p_1 - p_2 = 0 \rightarrow p_1 = p_2$ The alternative hypothesis is $H_1: p_1 - p_2 \neq 0 \rightarrow p_1 \neq p_2$ The value of the test statistic is z = -2.67.The p-value of the test is 0.0076 < 0.05(standard significance level), which means that there is enough evidence to conclude that the proportion of people who wear life vests while riding a jet ski is not the same as the proportion of people who wear life vests while riding in a boat.

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Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

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Proportion 1: Jet-ski users

86.5% out of 400, thus:

p_1 = 0.865

$s_1 = \sqrt{\frac{0.865*0.135}{400}} = 0.0171$

Proportion 2: boaters

92.8% out of 250, so:

p_2 = 0.928

$s_2 = \sqrt{\frac{0.928*0.072}{250}} = 0.0163$

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Hypothesis:

Test the claim that the proportion of people who wear life vests while riding a jet ski is not the same as the proportion of people who wear life vests while riding in a boat.

At the null hypothesis, it is tested that the proportions are the same, that is, the subtraction is 0. So

$H_0: p_1 - p_2 = 0 \rightarrow p_1 = p_2$

At the alternative hypothesis, it is tested that the proportions are different, that is, the subtraction is different of 0. So

$H_1: p_1 - p_2 \neq 0 \rightarrow p_1 \neq p_2$

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Test statistic:

The test statistic is:

$z = \frac{X - \mu}{s}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, and s is the standard error.

0 is tested at the null hypothesis.

This means that $\mu = 0$

From the samples:

X = p_1 - p_2 = 0.865 - 0.928 = -0.063

$s = \sqrt{s_1^2 + s_2^2} = \sqrt{0.0171^2 + 0.0163^2} = 0.0236$

The value of the test statistic is:

$z = \frac{X - \mu}{s}$

$z = \frac{-0.063 - 0}{0.0236}$

z = -2.67

The value of the test statistic is z = -2.67.

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p-value of the test and decision:

The p-value of the test is the probability that the proportion differs by at at least 0.063, which is P(|z| > 2.67), given by 2 multiplied by the p-value of z = -2.67.

Looking at the z-table, z = -2.67 has a p-value of 0.0038.

2*0.0038 = 0.0076.

The p-value of the test is 0.0076 < 0.05(standard significance level), which means that there is enough evidence to conclude that the proportion of people who wear life vests while riding a jet ski is not the same as the proportion of people who wear life vests while riding in a boat.

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