Question

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.

Answer 1

False

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 = (\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.

West University:

15 out of 100, so:

`p_W = (15)/(100) = 0.15`

`s_W = \sqrt{(0.15*0.85)/(100)} = 0.0357`

East University:

12 out of 100, so:

`p_E = (12)/(100) = 0.12`

`s_E = \sqrt{(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 \\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:

`X = p_W - p_E = 0.15 - 0.12 = 0.03`

`s = √(s_W^2+s_E^2) = √(0.0357^2+0.0325^2) = 0.0483`

Value of the test statistic:

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

`z = (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.