Determine if the proportion of lung cancer patients is different than the national average of 18%....

Determine if the proportion of lung cancer patients is different than the national average of 18%. Please use the p-value approach and α=0.05. (8 pts)
Construct and interpret the 95% confidence interval for the proportion of people with lung cancer. (4 pts)

Lung Cancer

No Lung Cancer

Total

Asbestos Exposure

153

201

354

No Asbestos Exposure

23

664

687

Total

176

865

1041

Expert Solution

(a)
Given that,
possible chances (x)=176
sample size(n)=1041
success rate ( p )= x/n = 0.1691
success probability,( po )=0.18
failure probability,( qo) = 0.82
null, Ho:p=0.18
alternate, H1: p!=0.18
level of significance, alpha = 0.05
from standard normal table, two tailed z alpha/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.16907-0.18/(sqrt(0.1476)/1041)
zo =-0.9181
| zo | =0.9181
critical value
the value of |z alpha| at los 0.05% is 1.96
we got |zo| =0.918 & | z alpha | =1.96
make decision
hence value of |zo | < | z alpha | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -0.91807 ) = 0.35858
hence value of p0.05 < 0.3586,here we do not reject Ho
------------------------------------------------------------------------------
null, Ho:p=0.18
alternate, H1: p!=0.18
test statistic: -0.9181
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.35858
no evidence that proportion of lung cancer patients is different than the national average of 18%
(b)
CI = confidence interval
confidence interval = [ 0.1691 ± 1.96 * Sqrt ( (0.1691*0.8309) /1041) ) ]
= [0.1691 - 1.96 * Sqrt ( (0.1691*0.8309) /1041) , 0.1691 + 1.96 * Sqrt ( (0.1691*0.8309) /1041) ]
= [0.1463 , 0.1918]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 95% sure that the interval [ 0.1463 , 0.1918] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion

orchestra answered 1 year ago

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