Answer: 0.228
Step-by-step explanation:
We know that the formula to find the upper limit of confidence interval for population proportion is given by :-
[tex]\hat{p}+ z*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex] , where
n= Sample size
[tex]\hat{p}[/tex] = Sample proportion
z* = critical value.
Let p be the proportion of customers who responds yes to a survey.
As per given , we have
n= 200
[tex]\hat{p}=\dfrac{35}{200}=0.175[/tex]
Confidence level : 95%
The critical z-value for 95% confidence is z* = 1.96 [ from z-table]
Substitute all values in the formula , we get
[tex]0.175+(1.96)\sqrt{\dfrac{0.175(1-0.175)}{200}}[/tex]
[tex]=0.175+(1.96)\sqrt{0.000721875}[/tex]
[tex]=0.175+(1.96)(0.0268677315753)[/tex]
[tex]=0.227660753888\approx0.228[/tex]
Hence, the upper limit of a 95% confidence level estimate of the population proportion is 0.228.
