Question

A publisher reports that 75% of their readers own a particular make of car. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 280 found that 70% of the readers owned a particular make of car. Is there sufficient evidence at the 0.02 level to support the executive's claim?

Answer 1

Answer:

The p-value of the test is of 0.0536 > 0.02, which means that there is not sufficient evidence at the 0.02 level to support the executive's claim.

Step-by-step explanation:

A publisher reports that 75% of their readers own a particular make of car. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage.

At the null hypothesis, we test if the proportion is of 75%, that is:

$H_0: p = 0.75$

At the alternate hypothesis, we test if the proportion is different of 75%, that is:

$H_1: p \neq 0.75$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

75% is tested at the null hypothesis:

This means that $\mu = 0.75, \sigma = \sqrt{0.75*0.25}$

A random sample of 280 found that 70% of the readers owned a particular make of car.

This means that $n = 280, X = 0.7$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.7 - 0.75}{\frac{\sqrt{0.75*0.25}}{\sqrt{280}}}$

$z = -1.93$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion differing from 0.75 by at least |0.7 - 0.75| = 0.05, which is P(|z| > 1.93), which is 2 multiplied by the p-value of z = -1.93.

Looking at the z-table, z = -1.93 has a p-value of 0.0268.

2*0.0268 = 0.0536.

The p-value of the test is of 0.0536 > 0.02, which means that there is not sufficient evidence at the 0.02 level to support the executive's claim.