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Esteffyy
27.03.2021 •
Mathematics
National statistics show that 23% of men smoke and 18.5% of women smoke. A random sample of 175 men indicated that 45 were smokers, and a random sample of 155 women surveyed indicated that 42 smoked. Construct a 98% confidence interval for the true difference in proportions of male and female smokers.
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Ответ:
98% C.I. = [-0.1270, 0.0994]
Step-by-step explanation:
Here
p1= proportion of men smokers
p1= 45/175= 0.2571
p2= proportion of women smokers
p2= 42/155= 0.2709
1) Choose the significance level ∝= 0.02
2) The formula for confidence interval is
(p1^- p2^)- z ∝/2sqrt( p1^(1-p1^)/n1 + ( p2^(1-p2^)/n2))
and
(p1^- p2^) z ∝/2sqrt( p1^(1-p1^)/n1 + ( p2^(1-p2^)/n2))
3) The z value is 2.05
4) Calculations:
(p1^- p2^)- z ∝/2sqrt( p1^(1-p1^)/n1 + ( p2^(1-p2^)/n2))
Putting the values
= 0.2571-0.2709)(-2.05) sqrt ( 0.2571(1-0.2571)/175+ 0.2709(1-0.2709)/155)
=-0.1270 ( using calculator)
And
(p1^- p2^)z ∝/2sqrt( p1^(1-p1^)/n1 + ( p2^(1-p2^)/n2))
Putting the values
= 0.2571-0.2709)(2.05) sqrt ( 0.2571(1-0.2571)/175+ 0.2709(1-0.2709)/155)
=0.0994 ( using calculator)
and Putting ± 2.05 in the confidence interval formula we get limits (-0.1270, 0.0994)
Ответ:
The answer is C
Step-by-step explanation: