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# Sample Size

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Having been retained by the National Native American Political Party to conduct a poll of residents of Muscogee, Oklahoma, you have worked diligently on completing interviews with residents of that town for three weeks. Your budget of \$10,000 is expended. You have completed 270 interviews.

(A) If you promised to deliver a survey that is accurate plus minus 5% at a confidence level of 90%, can you say, using a conservative basis for estimation, that you have reached your objective? (Assume normally distributed data.) If your answer is yes, justify. If no, is there any way of adjusting your procedure to meet the requirement?

(B) If the data were oriented toward simply determining whether the people of Muscogee would or would not vote for a bond issue, how would your analysis change? (Assume equal likelihood of yes versus no).

https://brainmass.com/statistics/confidence-interval/sample-size-calculate-accuracy-29690

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(A) If you promised to deliver a survey that is accurate plus minus 5% at a confidence level of 90%, can you say, using a conservative basis for estimation, that you have reached your objective? (Assume normally distributed data.) If your answer is yes, justify. If no, is there any way of adjusting your procedure to meet the requirement?

No of tails= 2
This is a 2 tailed test for normal distribution
as we are checking the accuracy of plus minus 5%

sp=standard error of proportion=square root of (pq/n)
(The maximum value of standard error of proportion for any given value of n is when p=q=0.5)

confidence interval= 90%
Z corresponding to 90% confidence interval and 2 tailed test is 1.6449
(from the tables or using EXCEL function NORMSINV)

We have to see that Z* sp < 5%
or sp < 5.%/Z or 5.% / 1.6449
or sp < 0.030397 =5%/1.6449
But
sp=standard error of proportion=square root of (pq/n)
or n=(pq)/(sp^2)
The maximum value of standard error of proportion for any given value of n is when p=q=0.5
We will use these values as a conservative basis for estimation
pq= 0.25 =0.5*0.5
or n=(pq)/(sp^2)= 270.57 =0.25/0.030397^2

Therefore sample size required for plus minus 5% accuracy is 270.57

Since we have completed 270 interviews we have reached our objective.

(B) If the data were oriented toward simply determining whether the people of Muscogee would or would not vote for a bond issue, how would your analysis change? (Assume equal likelihood of yes versus no).

Now we are checking on only one side
(We want to know whether p>0.5); if p>0.5 the people have voted

No of tails= 1
This is a 1 tailed test for normal distribution
as we are checking the accuracy of plus 5%

sp=standard error of proportion=square root of (pq/n)
(assume equal likelihood of yes versus no means p=q=0.5)

confidence interval= 90%
Z corresponding to 90% confidence interval and 1 tailed test is 1.2816
(from the tables or using EXCEL function NORMSINV)

We have to see that Z* sp < 5%
or sp < 5.%/Z or 5.% / 1.2816
or sp < 0.039014 =5%/1.2816
But
sp=standard error of proportion=square root of (pq/n)
or n=(pq)/(sp^2)

pq= 0.25 =0.5*0.5
or n=(pq)/(sp^2)= 165 =0.25/0.039014^2

Therefore sample size required is 165.

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