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.

Answers: 2

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Answers

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.

A similar question is found at

Answer from: Quest

not only in mathematics, scientific proof also are required in the other field of science. also maths proofs are of particular structure, containing steps and logics. so proving is like doing science, it is science. that's why it is important to prove.

an inaccurate proof leads to errors, or stating results that are not accurate. which wil mess the probelm up you dont want this

Answer from: Quest

2160ft3

step-by-step explanation: