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

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

Explanation:

The test statistic is:

`z = (X - \mu)/((\sigma)/(√(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 = √(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 = (411)/(900) = 0.4567`

Value of the test-statistic:

`z = (X - \mu)/((\sigma)/(√(n)))`

`z = (0.4567 - 0.5)/((0.5)/(√(900)))`

`z = -2.598`

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