Adults over the age of 18 were randomly selected to take part in a survey about

Question

Adults over the age of 18 were randomly selected to take part in a survey about if they are overweight to see if the rates differ by gender. Responses to then survey are coded in Excel as "Overweight" and "Healthy" with the responses from the males in Column A and the responses from the females in Column B. Construct a 99% confidence interval for the difference in the proportion of males and females who reported being overweight. (You may use Excel, but make sure you can do the calculations by hand.) a. b. Interpret the interval in the context of the problem. c. Does the data provide sufficient evidence to conclude that there is a significant difference in the proportion of males and females who are overweight? If so, determine which gender has a significantly higher proportion and explain why. If not, explain how you made that determination. Do the results of this study allow us to conclude that the difference in the average number of pounds that males d. and females are overweight by is significantly different? Briefly explain why or why not.

Explanation / Answer

a.
TRADITIONAL METHOD
given that,
sample one, x1 =953, n1 =2074, p1= x1/n1=0.4595
sample two, x2 =1097, n2 =2237, p2= x2/n2=0.4904
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.4595*0.5405/2074) +(0.4904 * 0.5096/2237))
=0.0152
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.58
margin of error = 2.58 * 0.0152
=0.0393
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.4595-0.4904) ±0.0393]
= [ -0.0701 , 0.0084]
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DIRECT METHOD
given that,
sample one, x1 =953, n1 =2074, p1= x1/n1=0.4595
sample two, x2 =1097, n2 =2237, p2= x2/n2=0.4904
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.4595-0.4904) ± 2.58 * 0.0152]
= [ -0.0701 , 0.0084 ]
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b.
interpretations:
1) we are 99% sure that the interval [ -0.0701 , 0.0084] contains the difference between
true population proportion P1-P2
2) if a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the difference between
true population mean P1-P2
c.
Given that,
sample one, x1 =953, n1 =2074, p1= x1/n1=0.459
sample two, x2 =1097, n2 =2237, p2= x2/n2=0.49
null, Ho: p1 = p2
alternate, H1: p1 < p2
level of significance, = 0.01
from standard normal table,left tailed z /2 =2.326
since our test is left-tailed
reject Ho, if zo < -2.326
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.459-0.49)/sqrt((0.476*0.524(1/2074+1/2237))
zo =-2.029
| zo | =2.029
critical value
the value of |z | at los 0.01% is 2.326
we got |zo| =2.029 & | z | =2.326
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: left tail - Ha : ( p < -2.0292 ) = 0.02122
hence value of p0.01 < 0.02122,here we do not reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 < p2
test statistic: -2.029
critical value: -2.326
decision: do not reject Ho
p-value: 0.02122
we do not have enough evidence to support the claim that female are overweight is higher proportion than males

d.
we conclude that the difference in average number of pounds that males and females are overweight by is significantly different because females proportion is higher than the males proportion
both are different proportion.

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