3.90 Teens Are More Likely to Send Text Messages Exercise 3.28 on page 210 compares studies which measure the proportions of adult and teen cell phone users thatsend/receive text messages. The summary statistics are repeated below:Group Sample Size ProportionPt 0.87Тeennt 800Adult na 2252 p = 0.72Figure 3.21 shows a distribution for the differences in sample proportions (p - pa) for 5000 bootstrap samples (taking 800 values with replacement from the original teensample and 2252 from the adults)Figure 3.21 Bootstrap difference in sample proportions of teen and adult cell phone users who text300250200150100500.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.20p_Teen p_Adulta. Based on the bootstrap distribution, which is the most reasonable estimate of the standard error for the difference in proportions: SE 0.015, 0.030, 0.050, 0.10, or 0.15?Explain the reason for your choiceb. Using your choice for the SE estimate in part (a), find and interpret a 95% confidence interval for the difference in proportion of teen and adult cell phone users whosend/receive text messages.Frequency

Question
Asked Oct 15, 2019

I'm working on study problems for an exam, and I'm absolutely stuck on this problem. 

3.90 Teens Are More Likely to Send Text Messages Exercise 3.28 on page 210 compares studies which measure the proportions of adult and teen cell phone users that
send/receive text messages. The summary statistics are repeated below:
Group Sample Size Proportion
Pt 0.87
Тeen
nt 800
Adult na 2252 p = 0.72
Figure 3.21 shows a distribution for the differences in sample proportions (p - pa) for 5000 bootstrap samples (taking 800 values with replacement from the original teen
sample and 2252 from the adults)
Figure 3.21 Bootstrap difference in sample proportions of teen and adult cell phone users who text
300
250
200
150
100
50
0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.20
p_Teen p_Adult
a. Based on the bootstrap distribution, which is the most reasonable estimate of the standard error for the difference in proportions: SE 0.015, 0.030, 0.050, 0.10, or 0.15?
Explain the reason for your choice
b. Using your choice for the SE estimate in part (a), find and interpret a 95% confidence interval for the difference in proportion of teen and adult cell phone users who
send/receive text messages.
Frequency
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3.90 Teens Are More Likely to Send Text Messages Exercise 3.28 on page 210 compares studies which measure the proportions of adult and teen cell phone users that send/receive text messages. The summary statistics are repeated below: Group Sample Size Proportion Pt 0.87 Тeen nt 800 Adult na 2252 p = 0.72 Figure 3.21 shows a distribution for the differences in sample proportions (p - pa) for 5000 bootstrap samples (taking 800 values with replacement from the original teen sample and 2252 from the adults) Figure 3.21 Bootstrap difference in sample proportions of teen and adult cell phone users who text 300 250 200 150 100 50 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.20 p_Teen p_Adult a. Based on the bootstrap distribution, which is the most reasonable estimate of the standard error for the difference in proportions: SE 0.015, 0.030, 0.050, 0.10, or 0.15? Explain the reason for your choice b. Using your choice for the SE estimate in part (a), find and interpret a 95% confidence interval for the difference in proportion of teen and adult cell phone users who send/receive text messages. Frequency

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Step 1

The formula for the estimate of the standard error for the difference in proportion is given below:

The formula for the 95% confidence interval for the difference in proportion is given below:

p.(1- р.) . Р.(1-Р.)
SE(P, -P.)
п,
п,
CI- (р. - Р.)+Z, SE (p. - P.)
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p.(1- р.) . Р.(1-Р.) SE(P, -P.) п, п, CI- (р. - Р.)+Z, SE (p. - P.)

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Step 2

a.

The estimate of the standard error for the ...

P. (1- Р.). Р.(1- Р.)
п,
SE(p.-Pa)
п,
a
0.87(1-0.87) 0.72(1-0.72)
800
2252
0.1131 0.2016
+
800
2252
=10.000141 0.00 0089
= V0.00023
=0.015
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P. (1- Р.). Р.(1- Р.) п, SE(p.-Pa) п, a 0.87(1-0.87) 0.72(1-0.72) 800 2252 0.1131 0.2016 + 800 2252 =10.000141 0.00 0089 = V0.00023 =0.015

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