Question

1. In a recent campus survey, 75 Indiana students were asked if they felt that their education at Cleary was preparing them for their future careers and 83% of students responded “Extremely well prepared.” Construct a 95% confidence interval (use z= 1.96) for the true proportion of Cleary students who feel the same way. Round standard error to 4 decimal places.

2.   Is there anything you could do to get a narrower range of values in the previous problem?

3. Is the sample size of 75 from the previous problem enough to get a 3% margin of error and 95% confidence? (To get full credit find the minimum sample size and then compare to 600 to see if large enough). Again use z=1.96. Also use a p-hat value of 0.83.

Answer 1

Question 1

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

n = 75

P = x/n = 0.83

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = .83 ± 1.96* sqrt(.83*(1 – .83)/75)

Confidence Interval = .83 ± 1.96* 0.0434

Confidence Interval = .83 ± 0.0850

Lower limit = .83 - 0.0850 = 0.7450

Upper limit = .83 + 0.0850 = 0.9150

Confidence interval = (0.7450, 0.9150)

2.   Is there anything you could do to get a narrower range of values in the previous problem?

Yes, we can decrease the sample size or confidence level to get a narrower range or width of the confidence interval.

3. Is the sample size of 75 from the previous problem enough to get a 3% margin of error and 95% confidence?

Confidence level = 95%

Z = 1.96

p = 0.83

q = 1 – p = 0.17

E = 0.03

n = p*q*(Z/E)^2

n = 0.83*0.17*(1.96/0.03)^2

n = 602.2775

Required sample size = 603

The sample size from previous problem is not enough to get required margin of error.