Question
Jan Villaroel
Topic: Statistics Posted 10 months ago
Participants in independent representative samples of teens and parents of teens were asked if they check their phone for messages

Participants in independent representative samples of teens and parents of teens were asked if they check their phone for messages as soon as they wake up. Of the 750 teens surveyed, 42% said "yes" and of the 1,050 parents of teens surveyed, 28% said "yes." Do the data provide convincing evidence that the proportion of parents of teens who check their phone for messages as soon as they wake up is less than the proportion of teens who check for messages as soon as they wake up? Test the relevant hypotheses using a significance level of 0.05. (Let p1 be the population proportion of parents of teens who check their phone for messages as soon as they wake up and p2 be the population proportion of teens who check their phone for messages as soon as they wake up.)

a) State the appropriate null and alternative hypotheses.

b) Find the test statistic. (Round your answer to two decimal places.)

c) Use technology to find the P-value. (Round your answer to four decimal places.)


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Koala Tutor
Feb 13, 2024
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a) The null and alternative hypotheses are:

Null hypothesis (H0): p1 = p2 (The proportion of parents of teens who check their phone for messages as soon as they wake up is equal to the proportion of teens who do the same)

Alternative hypothesis (Ha): p1 < p2 (The proportion of parents of teens who check their phone for messages as soon as they wake up is less than the proportion of teens who do the same)

b) To find the test statistic, we use the formula:

test statistic = (p1 - p2) / √((p1q1/n1) + (p2q2/n2))

Where:
p1 = proportion of parents of teens who check their phone for messages as soon as they wake up
p2 = proportion of teens who check their phone for messages as soon as they wake up
q1 = 1 - p1
q2 = 1 - p2
n1 = sample size of parents of teens
n2 = sample size of teens

Plugging in the values, we get:

test statistic = (0.28 - 0.42) / √((0.28*0.72/1050) + (0.42*0.58/750))

test statistic ≈ -5.080

c) Using technology (such as a calculator or statistical software), we find the P-value associated with a test statistic of -5.080.

P-value ≈ 0.0000

Since the P-value (0.0000) is less than the significance level (0.05), we reject the null hypothesis.

Therefore, the data provide convincing evidence that the proportion of parents of teens who check their phone for messages as soon as they wake up is less than the proportion of teens who do the same.

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