Question

A publisher reports that 35% of their readers own a laptop. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 190 found that 32% of the readers owned a laptop. Is there sufficient evidence at the 0.10 level to support the executive's claim

Answer 1

Answer:

The pvalue of the test is 0.3844 > 0.1, which means that there is not sufficient evidence at the 0.10 level to support the executive's claim

Step-by-step explanation:

A publisher reports that 35% of their readers own a laptop.

This means that the null hypothesis is:

$H_0: p = 0.35$

A marketing executive wants to test the claim that the percentage is actually different from the reported percentage.

This means that the alternate hypothesis is:

$H_{a}: p \neq 0.35$

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.

0.35 is tested at the null hypothesis:

This means that $\mu = 0.35, \sigma = \sqrt{0.35*0.65}$

A random sample of 190 found that 32% of the readers owned a laptop.

This means that $X = 0.32, n = 190$

Z-statistic:

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

$z = \frac{0.32 - 0.35}{\frac{\sqrt{0.35*0.65}}{\sqrt{190}}}$

$z = -0.87$

pvalue of the test and decision:

As we are testing that the mean is different from a value and z is negative, the pvalue of the test is 2 multiplied by the pvalue of z = -0.87

Looking at the z-table, z = -0.87 has a pvalue of 0.1922

2*0.1922 = 0.3844

0.3844 > 0.1, which means that there is not sufficient evidence at the 0.10 level to support the executive's claim