Question

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 500 randomly selected people who rode a jet ski, 85% wore life vests. Out of 250 randomly selected boaters, 90.4% wore life vests. Using a 0.05 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. State the null and alternative hypotheses for the test.

Answer 1

The null hypothesis is
`H_0: p_1 - p_2 = 0`

The alternate hypothesis is
`H_a: p_1 - p_2 \\eq 0`

The pvalue of the test is 0.03 < 0.05, which means that we have enough evidence to accept the alternative hypothesis that he 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.

Explanation:

Before testing the null hypothesis, we need to understand the Central Limit Theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes 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 = (\sigma)/(√(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{(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.

Out of 500 randomly selected people who rode a jet ski, 85% wore life vests.

This means that
`p_1 = 0.85, s_1 = \sqrt{(0.85*0.15)/(500)} = 0.016`

Out of 250 randomly selected boaters, 90.4% wore life vests.

This means that
`p_2 = 0.904, s_2 = \sqrt{\frac{{0.904*0.096}{250}} = 0.019`

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, we test that the proportions are the same, that is, the subtraction between them is zero. So

`H_0: p_1 - p_2 = 0`

At the alternate hypothesis, we test that the proportions are different, that is, the subtraction between them is different of zero. So

`H_a: p_1 - p_2 \\eq 0`

The test statistic is:

`z = (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 two samples, we have that:

`X = p_1 - p_2 = 0.85 - 0.904 = -0.054`

`s = √(s_1^2+s_2^2) = √(0.016^2 + 0.019^2) = 0.0248`

Test statistic:

`z = (X - \mu)/(s)`

`z = (-0.054 - 0)/(0.0248)`

`z = -2.17`

Pvalue of the test and decision:

The pvalue of the test is the probability that the difference differs from 0 by at least 0.054, which is P(|Z| > 2.17), which is 2 multiplied by the pvalue of z = -2.17

Looking at the z-table, z = -2.17 has a pvalue of 0.015

2*0.015 = 0.03

The pvalue of the test is 0.03 < 0.05, which means that we have enough evidence to accept the alternative hypothesis that he 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.