Question

A random sample of 900 13- to 17-year-olds found that 411 had a computer in their room with Internet access. Let p be the proportion of all teens in this age range who have a computer in their room with Internet access.Suppose you wished to see if the majority of teens in this age range have a computer in their room with Internet access. To do this, you test the hypotheses
H0: p = 0.50 vs HA : p≠0.50
The test statistic for this test is:_______.
a. 1.96
b. 2.60
c. 27.50
d. 2.59424

Answer 1

Answer:

The test statistic for this test is: z = -2.60, option b.

Step-by-step explanation:

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.

H0: p = 0.50 vs HA : p≠0.50

This means that $\mu = 0.5, \sigma = \sqrt{0.5*0.5} = 0.5$

A random sample of 900 13- to 17-year-olds found that 411 had a computer in their room with Internet access.

This means that $n = 900, X = \frac{411}{900} = 0.4567$

Value of the test-statistic:

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

$z = \frac{0.4567 - 0.5}{\frac{0.5}{\sqrt{900}}}$

$z = -2.598$

The test statistic for this test is: z = -2.60, option b(should be).