Question

Sarah collects data on members in team A and B and on proportions of students eye color.

	Brown eyes	Blue eyes
Team A	233	270
Team B	167	190

1. Construct an 88% confidence interval in eye color (order: Brown eyes in Team A - Brown eyes in Team B). Include all steps.

2. Do a hypothesis test (p-values) to see whether the proportions of brown color is the same for Team A and B. alpha level = 0.05. Include all steps.

Answer 1

1)

N1 = 233+270 = 503, P1 = 233/503

N2 = 167 + 190 = 357, P2 = 167/357

First we need to check the conditions of normality that is if n1p1 and n1*(1-p1) and n2*p2 and n2*(1-p2) all are greater than equal to 5 or not

N1*p1 = 233

N1*(1-p1) = 270

N2*p2 = 167

N2*(1-p2) = 190

All the conditions are met so we can use standard normal z table to estimate the interval.

Critical value z from z table for 88% interval is 1.55

Margin of error (MOE) = Z*√{P1*(1-P1)/N1 + P2*(1-P2)/N2}

Interval is given by,

(P1-P2)- MOE < (P1-P2) < (P1-P2)+MOE

=  (-0.058, 0.049)

B)

Null hypothesis Ho : P1 = P2

Alternate hypothesis Ha : P1 not equal to P2

First we need to check the conditions of normality that is if n1p1 and n1*(1-p1) and n2*p2 and n2*(1-p2) all are greater than equal to 5 or not

N1*p1 = 233

N1*(1-p1) = 270

N2*p2 = 167

N2*(1-p2) = 190

All the conditions are met so we can use standard normal z table to conduct the test

Test statistics z = (P1-P2)/standard error

Standard error = √{p*(1-p)}*√{(1/n1)+(1/n2)}

P = pooled proportion = [(p1*n1)+(p2*n2)]/[n1+n2]

After substitution

Test statistics z = -0.13

From z table, P(Z<-0.13) = 0.44828

As the test is two tailed,

So, P-Value = 2*0.44828 = 0.89656

As the obtained P-Value is greater than the given significance.

We fail to reject the null hypothesis Ho.

We have enough evidence to conclude that the proportions of brown color is the same for Team A and B.