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
Question
Answer:
The test statistic for this test is: z = -2.60, option b.
Step-by-step explanation:
The test statistic is:
[tex]z = \frac{X – \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
H0: p = 0.50 vs HA : p≠0.50
This means that [tex]\mu = 0.5, \sigma = \sqrt{0.5*0.5} = 0.5[/tex]
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 [tex]n = 900, X = \frac{411}{900} = 0.4567[/tex]
Value of the test-statistic:
[tex]z = \frac{X – \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.4567 – 0.5}{\frac{0.5}{\sqrt{900}}}[/tex]
[tex]z = -2.598[/tex]
The test statistic for this test is: z = -2.60, option b(should be).