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

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

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bonexptip 4 years 2021-08-27T15:28:54+00:00 1 Answers 8 views 0

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  1. taiduc
    0
    2021-08-27T15:29:54+00:00
    Reply

    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).

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