Question

n 200 5. [12 marks] Workers in several industries were surveyed to determine the proportion of workers who feel their industry is understaffed. In the government sector, 37% of the respondents said they were understaffed, in the health care sector 33% said they were understaffed and in the education sector 28% said they were understaffed (USA Today, January 11, 2010). Suppose that 200 workers were surveyed in each industry. 16-2 each] Construct a 95 % confidence interval for the proportion of workers in each of a) these industries who feel their industry is understaffed. 7% Government: -P 2 PL-P> 031 196 0.57U-051 00 (0.30. 0.44 Health care: 37 P2o.33-s3) 203 19633U-0.55 , 0.40) Education: =PZPU-P) 0.2S 1.9625C1-a) -ro.22, 0.34) TOTAL marks for page [6 Page 7 of 10

Answer 1

Confidence interval for proportion

$\hat{p} \pm Zcritical * \sqrt{\frac{\hat{p}(1-\hat{p})}{n} }$

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(a) For Government

$= 0.37 \pm 1.96 * \sqrt{\frac{0.37*0.63}{200}}$

Lower limit = 0.37 - 0.067 = 0.303

Upper Limit = 0.37 + 0.067 = 0.437

The Confidence interval is (0.303, 0.437)

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For Health Care

$= 0.33 \pm 1.96 * \sqrt{\frac{0.33*0.67}{200}}$

Lower limit = 0.33 - 0.065 = 0.265

Upper Limit = 0.33 + 0.065 = 0.395

The Confidence interval is (0.265, 0.395)

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For Education

$= 0.28 \pm 1.96 * \sqrt{\frac{0.28*0.72}{200}}$

Lower limit = 0.28 - 0.062 = 0.218

Upper Limit = 0.28 + 0.062 = 0.342

The Confidence interval is 0.218, 0.342

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SAMPLE SIZE

The ME = Zcritical * SQRT(p * q/n).

Squaring and solving, n = (Zc/ME)² * p * q

For Government

Pilot Study

p = 0.37, q = 0.63, ME = 0.05

n = 358.19 $\approx$    359
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For Health Care

Pilot Study

p = 0.33, q = 0.67, ME = 0.05

n = 339.75 $\approx$    340

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For Education

Pilot Study

p = 0.28, q = 0.67, ME = 0.05

n = 309.79 $\approx$    310

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Worst Case Scenario (For all three)

p = 0.5, q = 0.5, ME = 0.05

n = 384.16 $\approx$    385

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