91 a In a random sample of 500 observations we found the pro

9.1.

a. In a random sample of 500 observations, we found the proportion of successes to be 48%. Estimate with 95% confidence the population proportion of successes.

b. Repeat part (a) with .

c. Repeat part (a) with .

d. Describe the effect on the confidence interval estimate of increasing the sample size.

Solution

a)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.48          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.022342784          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
Margin of error = z(alpha/2)*sp =    0.043791052          
lower bound = p^ - z(alpha/2) * sp =   0.436208948          
upper bound = p^ + z(alpha/2) * sp =    0.523791052          
              
Thus, the confidence interval is              
              
(   0.436208948   ,   0.523791052   ) [ANSWER]

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Hi! Some symbols are missing in parts b and c. Please resubmit this question together with the missing symbols there. Thanks!

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d)

As we increase the sample size, the standard error of the proportion decreases, so the confidence interval becomes narrower. [ANSWER]