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Normal distribution

1) Assume that the population proportions are to be estimated from the samples described. Find the margin of error and the 95% confidence interval.

Sample size = 256, sample proportion = 0.6

2) Importance: A Phone survey of 500 people revealed that 99% of those surveyed ranked good relationships as important or very important, 98% ranked financial security as important or very important. Third on the list was religious fulfillment at 86%, followed by a good sex life at 82%, and job satisfaction at 79%. The margin of error for the study was reported at 4% points.

Is the reported margin of error consistent with the sample size?

3) For each of the following, give the 95% confidence interval:

a) 17.2% of people in Florida are inactive, sample size = 1,500.
b) 13.2% of people in California have poor diets, sample size = 2,500.
c) 22.9% of people in Hawaii eat pineapple regularly, sample size = 3,500

4) A random selection of 1,600 people found that 900 support a certain lawyer in a political race. Based on the sample, would you claim that the lawyer will win a majority of the votes?

5) Estimate the minimum sample size needed to achieve the following margins of error:

E = 0.02, E = 0.05, E = 0.04, E = 0.035

6) Margin of error: In general, if one wished to decrease the margin or error by a factor of 2 in an estimation of a population proportion from E = 0.02 to 0.01, by what factor should the sample size be increased?

Solution Preview

Assume that the population proportions are to be estimated from the samples described. Find the margin of error and the 95% confidence interval.

Sample size = 256, sample proportion = 0.6

p= 0.60
q=1-p= 0.40
n=sample size= 256
sp=standard error of proportion=square root of (pq/n)= 0.0306 =square root of ( .6 * .4 / 256)
Confidence level= 95%
Significance level=alpha (a) = 5% =100% -95%
No of tails= 2
This is 2 tailed because we are calculating upper and lower confidence level

Since sample size= 256 >= 30 use normal distribution
Z at the 0.05 level of significance 2 tailed test = 1.96

Margin of error = + or - (z*sp)= 0.06 =1.96*0.0306

Upper confidence limit= p+z*sp= 0.66 =.6+1.96*.0306
Lower confidence limit= p-z*sp= 0.54 =.6-1.96*.0306

Answer:
Margin of error= 0.06
Upper confidence limit= 0.66
Lower confidence limit= 0.54

Importance:

"A Phone survey of 500 people revealed that 99% of those surveyed ranked good relationships as important or very important, 98% ranked financial security as important or very important. Third on the list was religious fulfillment at 86%, followed by a good sex life at 82%, and job satisfaction at 79%. The margin of error for the study was reported at 4% points.
Is the reported margin of error consistent with the sample size?"

p q pq
99% 1% 0.00990
98% 2% 0.01960
86% 14% 0.12040
82% 18% 0.14760
79% 21% 0.16590

The maximum value of pq corresponds to job satisfaction p=79%
we will take this value of pq for our calculations

The confidence interval has not been mentioned
We take it to be 95%

No of tails= 2
This is a 2 tailed test
as we are checking the accuracy of plus minus 4%

sp=standard error of proportion=square root of (pq/n)

confidence interval= 95%
Z corresponding to 95% confidence interval and 2 tailed test is 1.96
(from the tables or using EXCEL function NORMSINV)

We have to see that Z* sp < 4%
or sp < 4.%/Z or 4.% / 1.96
or sp < 0.020408 =4%/1.96
But
sp=standard error of proportion=square root of (pq/n)
or n=(pq)/(sp^2)

pq= 0.16590 =0.79*0.21
or n=(pq)/(sp^2)= 399 =0.1659/0.020408^2

Therefore sample size required for plus minus 4% ...

Solution Summary

Answers questions on estimation of proportions, confidence interval, margin of error, minimum sample size etc.

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