Question

Part 1) (a) Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the proportions of boys and girls under 10 years old who are afraid of spiders. Assume the "worst case" scenario for the value of both sample proportions. We want a 9999% confidence level and for the error to be smaller than 0.08.0.08.

Answer:

(b) Again find the sample size required, as in part (a), but with the knowledge that a similar student last year found that the proportion of boys afraid of spiders is 0.64 and the proportion of girls afraid of spiders was 0.54.

Answer:

Part 2)

Two random samples are taken, one from among UVA students and the other from among UNC students. Both groups are asked if academics are their top priority. A summary of the sample sizes and proportions of each group answering yes'' are given below:

UVA (Pop. 1):UNC (Pop. 2):n1=81,n2=90,p̂ 1=0.83p̂ 2=0.555UVA (Pop. 1):n1=81,p^1=0.83UNC (Pop. 2):n2=90,p^2=0.555

Find a 93.3% confidence interval for the difference p1−p2p1−p2 of the population proportions.

Confidence interval =

Part 3)

Some shrubs have the useful ability to resprout from their roots after their tops are destroyed. Fire is a particular threat to shrubs in dry climates, as it can injure the roots as well as destroy the aboveground material. One study of resprouting took place in a dry area of Mexico. The investigation clipped the tops of samples of several species of shrubs. In some cases, they also applied a propane torch to the stumps to simulate a fire. Of 20 specimens of a particular species, 4 resprouted after fire. Estimate with 99.8% confidence the proportion of all shrubs of this species that will resprout after fire.

Interval: ______to ______

Answer 1

Part 1. a)

Proportion for sample 1, p̂₁ = 0.5  

Proportion for sample 2, p̂₂ = 0.5

Margin of error, E = 0.08

Confidence Level, CL = 0.99

Significance level, α = 1 - CL = 0.01

Critical value, z = NORM.S.INV(0.01/2) = 2.5758

Sample size, n = (z² * (p̂₁*(1-p̂₁) + p̂₂*(1-p̂₂)) / E²

= 518.35 = 518

b)

Proportion for sample 1, p̂₁ = 0.64  

Proportion for sample 2, p̂₂ = 0.54

Margin of error, E = 0.08

Confidence Level, CL = 0.99

Significance level, α = 1 - CL = 0.01

Critical value, z = NORM.S.INV(0.01/2) = 2.5758

Sample size, n = (z² * (p̂₁*(1-p̂₁) + p̂₂*(1-p̂₂)) / E²

= 496.37 = 496

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Part 2.

For Sample 1: n1 = 81, p̂1 = 0.83

For Sample 2: n2 = 90, p̂2 = 0.555

93.3% Confidence interval for the difference:

At α = 0.067, two tailed critical value, z_c = NORM.S.INV(0.067/2) = 1.832

Lower Bound = (p̂1 - p̂2) - z_c*√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ]

= (0.83 - 0.555) - 1.832*√[(0.83*0.17/81) + (0.555*0.445/90)] = 0.152

Upper Bound = (p̂1 - p̂2) + z_c*√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ]

= (0.83 - 0.555) + 1.832*√[(0.83*0.17/81) + (0.555*0.445/90)] = 0.398

0.152 < p1 -p2 < 0.398

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Part 3.

n = 20, x = 4

p֮ = (x+2)/(n+4) = 0.25

Plus four 99.8% Confidence interval :

At α = 0.002, two tailed critical value, z_c = NORM.S.INV(0.002/2) = 3.090

Lower Bound = p֮ - z_c*√(p֮ *(1-p֮)/(n+4)) = 0.25 - 3.09 *√(0.25*0.75/(20+4)) = -0.0231

Upper Bound = p֮ + z_c*√(p֮ *(1-p֮ )/(n+4)) = 0.25 + 3.09 *√(0.25*0.75/(20+4)) = 0.5231

-0.0231 < p < 0.5231