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# Normal distribution

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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?

https://brainmass.com/statistics/confidence-interval/normal-distribution-41096

#### 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

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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