Question

A) You want to estimate what percent of a school has a dog so you survey 50 people. 20 tell you they have a dog at home. What is your best guess at the true proportion of people who have a dog at home? What is your 95% confidence interval? Your 90% confidence interval? How many people would you need to survey to have a 95% confidence interval of 5% in length.

B) You ask the 20 people with dogs how much their dog weighs. You get a mean response of 30lbs with a standard deviation of 8lbs. What is your 95% confidence interval for the average weight of the school student's dog?

Answer 1

Part a)

p̂ = X / n = 20/50 = 0.4
p̂ ± Z(α/2) √( (p * q) / n)
0.4 ± Z(0.05/2) √( (0.4 * 0.6) / 50)
Z(α/2) = Z(0.05/2) = 1.96
Lower Limit = 0.4 - Z(0.05) √( (0.4 * 0.6) / 50) = 0.2642
upper Limit = 0.4 + Z(0.05) √( (0.4 * 0.6) / 50) = 0.5358
95% Confidence interval is ( 0.2642 , 0.5358 )
( 0.2642 < P < 0.5358 )

p̂ = X / n = 20/50 = 0.4
p̂ ± Z(α/2) √( (p * q) / n)
0.4 ± Z(0.1/2) √( (0.4 * 0.6) / 50)
Z(α/2) = Z(0.1/2) = 1.645
Lower Limit = 0.4 - Z(0.1) √( (0.4 * 0.6) / 50) = 0.2860
upper Limit = 0.4 + Z(0.1) √( (0.4 * 0.6) / 50) = 0.5140
90% Confidence interval is ( 0.286 , 0.514 )
( 0.2860 < P < 0.5140 )

p̂ = 0.4
q̂ = 1 - p̂ = 0.6
Critical value Z(α/2) = Z(0.05/2) = 1.96 ( From Z table )
n = ( Z(α/2)² * p̂ * q̂ )/e²
n = ( Z(0.05)² * 0.4 * 0.6)/ 0.05²
n = 369
Required sample size at 95% confident is 369.

Part b)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 20- 1 ) = 2.093 ( Critical value from t table )
30 ± t(0.05/2, 20 -1) * 8/√(20)
Lower Limit = 30 - t(0.05/2, 20 -1) 8/√(20)
Lower Limit = 26.2559
Upper Limit = 30 + t(0.05/2, 20 -1) 8/√(20)
Upper Limit = 33.7441
95% Confidence interval is ( 26.2559 , 33.7441 )