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mustafajibawi1
05.08.2019 •
Mathematics
In a random sample of 300 women, 49% favored stricter gun control legislation. in a random sample of 200 men, 28% favored stricter gun control legislation. construct a 98% confidence interval for the difference between the population proportions p1 - p2
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Ответ:
Construct a 98% confidence interval for the difference between the population proportions p1 - p2 is ( 0.110199 and 0.309801)
Step-by-step explanation:
Given data
sample n1 = 300 women
sample n2 = 200 men
control legislation f1 = 49% = 0.49
control legislation f2 = 28% = 0.28
to find out
Construct a 98% confidence interval for the difference between the population proportions p1 - p2
solution
we know difference of control legislation = f1 -f2
so difference is 0.49-0.28 = 0.21
so for 98% confidence Z value is 2.326 from standard table
and interval for difference formula is
f1 - f2 - Z![\sqrt{f1(1-f1)/n1 + f2(1-f2)/n2}](/tpl/images/0171/9097/cfb98.png)
and
f1 - f2 + Z![\sqrt{f1(1-f1)/n1 + f2(1-f2)/n2}](/tpl/images/0171/9097/cfb98.png)
put all these value and we get difference of population proportion are
0.49 - 0.28 - 2.326![\sqrt{0.49(1-0.49)/300 + 0.28(1-0.28)/200}](/tpl/images/0171/9097/0956a.png)
and
0.49 - 0.28 + 2.326![\sqrt{0.49(1-0.49)/300 + 0.28(1-0.28)/200}](/tpl/images/0171/9097/0956a.png)
= 0.110199
and
= 0.309801
Construct a 98% confidence interval for the difference between the population proportions p1 - p2 is ( 0.110199 and 0.309801)
Ответ: