Question

The following data was obtained from two independent samples for which sigma was unknown; they were selected from two normally distributed populations.

Sample 1: 27, 39, 25, 33, 21, 35, 30, 26, 25, 31, 35, 30, 28

Sample 2: 24, 28, 23, 25, 24, 22, 29, 26, 29, 28, 19, 29

1. What is the point estimate for μ1- μ2?

2. Construct a 98% confidence interval for μ1- μ2.

3. Test at the 1% significance level whether the population mean for the first

   sample is greater than the population mean of the second sample.

Answer 1

Solution:

A point estimate for the difference in two population means is the difference in the corresponding sample means.

1)Point estimate of mu1-mu2= x1bar-x2bar

= 29.62- 25.5 = 4.12

Sample 1:

Sample 2:

2)98% confidence interval for μ1- μ2:

C. I = ((mu1-mu2) +- t(0.02) for n1+n2-2 d.f*S. E( mu1- mu2) )

= (4.12- 2.18 *1.7 , 4.12+2.18*1.7) = (0.414,7.826)

3)Testing at the 1% significance level :

Sample 1:

N1: 13

df1 = N - 1 = 13 - 1 = 12

M1: 29.62

SS1: 299.08

s21 = SS1/(N - 1) = 299.08/(13-1) = 24.92

Sample 2

N2: 12

df2 = N - 1 = 12 - 1 = 11

M2: 25.5

SS2: 115

s22 = SS2/(N - 1) = 115/(12-1) = 10.45

T-value Calculation:

s2p = ((df1/(df1 + df2)) * s21) + ((df2/(df2 + df2)) * s22) = ((12/23) * 24.92) + ((11/23) * 10.45) = 18

s2M1 = s2p/N1 = 18/13 = 1.38

s2M2 = s2p/N2 = 18/12 = 1.5

t = (M1 - M2)/√(s2M1 + s2M2) = 4.12/√2.89 = 2.42

The t-value is 2.42285. The p-value is .011839. The result is not significant at p < .01.

Hence we reject the Null hypothesis and accept the claim at 1% significance level that the population mean for the first sample is greater than the population mean of the second sample.