Mathematics, 20.07.2021 01:00 millie54

A study was conducted to determine if there was a difference in the driving ability of students from West University and East University by sending a survey to a sample of 100 students at both universities. Of the 100 sampled from West University, 15 reported they were involved in a car accident within the past year. Of the 100 randomly sampled students from East University, 12 students reported they were involved in a car accident within the past year. True or False. The difference in driving abilities at the two universities is statistically significant at the .05 significance level.

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Answer from: Dashavu4626

False

Step-by-step explanation:

Before testing the hypothesis, we need to understand the central limit theorem and subtraction of normal variables.

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.

West University:

15 out of 100, so:

$p_W = \frac{15}{100} = 0.15$

$s_W = \sqrt{\frac{0.15*0.85}{100}} = 0.0357$

East University:

12 out of 100, so:

$p_E = \frac{12}{100} = 0.12$

$s_E = \sqrt{\frac{0.12*0.88}{100}} = 0.0325$

Test the difference in driving abilities at the two universities:

At the null hypothesis we test if there is no difference, that is, the subtraction of the proportions is 0, so:

H_0: p_W - p_E = 0

At the alternative hypothesis, we test if there is a difference, that is, if the subtraction of the proportions is different of 0. So

$H_1: p_W - p_E \neq 0$

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 two samples:

X = p_W - p_E = 0.15 - 0.12 = 0.03

$s = \sqrt{s_W^2+s_E^2} = \sqrt{0.0357^2+0.0325^2} = 0.0483$

Value of the test statistic:

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

$z = \frac{0.03 - 0}{0.0483}$

z = 0.62

P-value of the test and decision:

The p-value of the test is the probability that the proportions differ by at least 0.03, which is P(|z| > 0.62), that is, 2 multiplied by the p-value of z = -0.62.

Looking at the z-table, z = -0.62 has a p-value of 0.2676.

2*0.2676 = 0.5352.

The p-value of the test is 0.5352 > 0.05, which means that the difference in driving is not statistically significant at the .05 significance level, and thus the answer is False.

Answer from: Quest

The united states declared war on japan the same exact day they attacked pearl harbor, which caused the death of many americans.

Answer from: Quest

2 is the final ! : )

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