#19) Complete parts a & b A recent survey of 1000 American women between the age
ID: 3054518 • Letter: #
Question
#19) Complete parts a & b
A recent survey of 1000 American women between the ages of 45 and 64 asked them what medical condition they most feared. Of those sampled, 63% said breast cancer, 6% said heart disease, and the rest picked other conditions. By contrast, currently about 3% of female deaths are due to breast cancer, whereas 32% are due to heart disease. Complete parts a and b below. a. Construct a 90% confidence interval for the population proportion of women who most feared breast cancer. Interpret. Vfalls from to We can be | % confident that the population proportion of women who Type integers or decimals rounded to three decimal places as needed.)Explanation / Answer
TRADITIONAL METHOD
given that,
possibile chances (x)= 63% * 1000 = 630
sample size(n)=1000
success rate ( p )= x/n = 0.63
I.
sample proportion = 0.63
standard error = Sqrt ( (0.63*0.37) /1000) )
= 0.0153
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, ? = 0.1
from standard normal table, two tailed z ?/2 =1.645
margin of error = 1.645 * 0.0153
= 0.0251
III.
CI = [ p ± margin of error ]
confidence interval = [0.63 ± 0.0251]
= [ 0.6049 , 0.6551]
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DIRECT METHOD
given that,
possibile chances (x)=630
sample size(n)=1000
success rate ( p )= x/n = 0.63
CI = confidence interval
confidence interval = [ 0.63 ± 1.645 * Sqrt ( (0.63*0.37) /1000) ) ]
= [0.63 - 1.645 * Sqrt ( (0.63*0.37) /1000) , 0.63 + 1.645 * Sqrt ( (0.63*0.37) /1000) ]
= [0.6049 , 0.6551]
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interpretations:
1. We are 90% sure that the interval [ 0.6049 , 0.6551] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion
roud off to 3 decimals [0.605 , 0.655]
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