An online site presented this question Would the recent noro

An online site presented this question, \'Would the recent norovirus outbreak deter you from taking a cruise?\' Among the 34,514 people who responded, 61% answered \'yes\'. Use the sample data to construct a 95% confidence interval estimate for the proportion of the population of all people who would respond \'yes\' to that question. Does the confidence interval provide a good estimate of the population proportion?

_____< P < _____

Does the confidence interval provide a good estimate of the population proportion?

Yes, the sample is large enough to provide a good estimate of the population proportion.

No, the responses are not independent.

Yes, all the assumptions for a confidence interval are satisfied.

No, the sample is a voluntary sample and might not be representative of the population.

Solution

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.61          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.002625425          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    2.33          
Thus,              
              
lower bound = p^ - z(alpha/2) * sp =   0.60388276          
upper bound = p^ + z(alpha/2) * sp =    0.61611724          
              
Thus, the confidence interval is              
              
(   0.60388276   ,   0.61611724   ) [ANSWER]

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Yes, the sample is large enough to provide a good estimate of the population proportion.

[As we can see, the confidence interval is already too narrow.]