Question

Q16. In a sample of 50 students, 60% lived at home. Estimate the true proportion of student who live at home using a confidence level of 90%. Conclude your analysis with a sentence.

Answer 1

n = 50,

60% of 50 = 30

$\hat{p}$ = 30 / 50 = 0.60

$\hat{q}$ = 1 -  $\hat{p}$ = 1 - 0.60 = 0.40

Confidence level = 90% = 0.90

α = 1 - 0.90 = 0.10

P value = 1 - (α/2)

= 1 - (0.10/2)

= 1 - 0.05

= 0.95

From z score table;

Zα_/2 = 1.65

Lower limit =  $\hat{p}$ - Zα_/2  √ ( $\hat{p}$ $\hat{q}$ / n)

= 0.60 - 1.65 √ [ (0.60)(0.40)/50 ]

= 0.60 - 1.65 √ 0.0048

= 0.60 - 0.1143

= 0.4857  

Upper limit =  $\hat{p}$   + Zα_/2  √ ( $\hat{p}$ $\hat{q}$ / n)

= 0.60 + 1.65 √ [ (0.60)(0.40)/50 ]

= 0.60 + 1.65 √ 0.0048

= 0.60 +  0.1143

= 0.7143

Conclusion:

0.4857 < P < 0.7143

"We are 90% confidence that the population of the students lived at home is between 48.57% and 71.43%"