Question

A study was conducted to compare the proportion of drivers in Boston and New York who wore seat belts while driving. Data were collected, and the proportion wearing seat belts in Boston was 0.581 and the proportion wearing seat belts in New York was 0.832. Due to local laws at the time the study was conducted, it was suspected that a smaller proportion of drivers wear seat belts in Boston than New York.
Required:
a. Find the test statistic for this test using Ha: pB < pNY. (Use standard error = 0.1.)
b. Determine the p value.

Answer 1

Answer:

a) The test statistic is $z = -2.51$

b) The pvalue is 0.0060.

Step-by-step explanation:

Question a:

Ha: pB < pNY

This means that the alternate hypothesis is rewritten as:

$H_a: p_{B} - p_{NY} < 0$

While the null hypothesis is:

$H_0: p_{B} - p_{NY} = 0$

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

pB = 0.581, pNY = 0.832

This means that $X = 0.581 - 0.832 = -0.251$

0 is tested at the null hypothesis:

This means that $\mu = 0$

Standard error = 0.1

This means that $s = 0.1$

Test statistic:

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

$z = \frac{-0.251 - 0}{0.1}$

$z = -2.51$

The test statistic is $z = -2.51$

Question b:

The pvalue is the probability of finding a proportion less than -0.251, that is, a difference of at least 0.251, which is the pvalue of Z = -2.51.

Looking at the z-table, Z = -2.51 has a pvalue of 0.0060

The pvalue is 0.0060.