Question

Consider the following data drawn independently from normally distributed populations: (You may find it useful to reference the appropriate table: z table or t table)

x−1x−1 = −17.7	x−2x−2 = −16.4
s12 = 8.0	s22 = 8.4
n1 = 20	n2 = 29

a. Construct the 95% confidence interval for the difference between the population means. Assume the population variances are unknown but equal. (Round all intermediate calculations to at least 4 decimal places and final answers to 2 decimal places.)
  

b. Specify the competing hypotheses in order to determine whether or not the population means differ.
  

• H₀: μ₁ − μ₂ = 0; H_A: μ₁ − μ₂ ≠ 0

• H₀: μ₁ − μ₂ ≥ 0; H_A: μ₁ − μ₂ < 0

• H₀: μ₁ − μ₂ ≤ 0; H_A: μ₁ − μ₂ > 0

c. Using the confidence interval from part a, can you reject the null hypothesis?
  

• Yes, since the confidence interval includes the hypothesized value of 0.

• Yes, since the confidence interval does not include the hypothesized value of 0.

• No, since the confidence interval includes the hypothesized value of 0.

• No, since the confidence interval does not include the hypothesized value of 0.

d. Interpret the results at   αα = 0.05.

• We cannot conclude that population mean 1 is greater than population mean 2.

• We conclude that population mean 1 is greater than population mean 2.

• We cannot conclude that the population means differ.

• We conclude that the population means differ

Answer 1

Solution:

Part a

Confidence interval for difference between two population means is given as below:

Confidence interval = (X1bar – X2bar) ± t*sqrt[Sp2*((1/n1)+(1/n2))]

Where Sp2 is pooled variance

Sp2 = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

We are given

X1bar = -17.7

X2bar = -16.4

S1^2 = 8

S2^2 = 8.4

n1 = 20

n2 = 29

df = n1 + n2 – 2 = 20 + 29 – 2 = 47

Confidence level = 95%

Critical value t = 2.0117

(by using t-table)

Sp2 = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

Sp2 = [(20 – 1)*8 + (29 – 1)*8.4]/(20 + 29 – 2)

Sp2 = 8.2383

Confidence interval = (X1bar – X2bar) ± t*sqrt[Sp2*((1/n1)+(1/n2))]

Confidence interval = ((-17.7) – (-16.4)) ± 2.0117*sqrt[8.2383*((1/20)+(1/29))]

Confidence interval = -1.3 ± 2.0117*0.8343

Confidence interval = -1.3 ± 1.6783

Lower limit = -1.3 - 1.6783 = -2.9783

Upper limit = -1.3 + 1.6783 = 0.3783

Confidence interval = (-2.98, 0.38)

Part b

H0: μ1 − μ2 = 0; HA: μ1 − μ2 ≠ 0

[Alternative hypothesis is ‘not equal to’, because we want to check whether two means are different or not.]

Part c

No, since the confidence interval includes the hypothesized value of 0.

Part d

We cannot conclude that the population means differ.