Question

1) A friend has performed a significance test of the null hypothesis that two means are equal. His report states that the null hypothesis is rejected in favor of the alternative that the first mean is larger than the second. In a presentation on his work, he notes that the first sample mean was larger than the second mean and this is why he chose this particular one-sided alternative. Suppose he reported t = 1.90 with a P-value of 0.04. What is the correct P-value that he should report?

2) A study of iron deficiency among infants compared samples of infants following different feeding regimens. One group contained breast-fed infants, while the infants in another group were fed a standard baby formula without any iron supplements. Here are summary results on blood hemoglobin levels at 12 months of age.

Group	n	x	s
Breast-fed	23	13.1	1.7
Formula	18	12.6	1.8

(a) Carry out a t test. Give the P-value. (Use α = 0.01. Use μ_breast-fed − μ_formula. Round your value for t to three decimal places, and round your P-value to four decimal places.)

t	=
P-value	=

(b) Give a 95% confidence interval for the mean difference in hemoglobin level between the two populations of infants. (Round your answers to three decimal places.)

Answer 1

Question 1

P-value for one tailed test = 0.04

So, P-value for two tailed test = 2*0.04 = 0.08

Correct P-value = 0.08

Question 2

Part a

Here, we have to use two sample t test for the difference between two population means assuming equal population variances.

H₀: µ₁ = µ₂ versus H_a: µ₁ ≠ µ₂

Test statistic formula for pooled variance t test is given as below:

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

We are given

X1bar = 13.1

X2bar = 12.6

S1 = 1.7

S2 = 1.8

n1 = 23

n2 = 18

df = n1 + n2 – 2 = 23 + 18 – 2 = 39

α = 0.01

Critical value = -2.7079 and 2.7079

(by using t-table)

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = [(23 – 1)*1.7^2 + (18 – 1)*1.8^2]/(23 + 18 – 2)

S_p² = 3.0426

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

t = (13.1 – 12.6) / sqrt[3.0426*((1/23)+(1/18))]

t = 0.5 / 0.5489

t = 0.9109

Test statistic = t = 0.911

P-value = 0.3679

P-value > α = 0.01

So, we do not reject the null hypothesis

Part b

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

Confidence interval = (X1bar – X2bar) ± t*sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = 3.0426

(X1bar – X2bar) = (13.1 – 12.6) = 0.5

Confidence level = 95%

df = 39

t = 2.0227

(by using t-table)

Standard error = sqrt[S_p²*((1/n1)+(1/n2))] = sqrt[3.0426*((1/23)+(1/18))] = 0.5489

Margin of error = t*sqrt[S_p²*((1/n1)+(1/n2))] = 2.0227*0.5489 = 1.1103

Confidence interval = (X1bar – X2bar) ± t*sqrt[S_p²*((1/n1)+(1/n2))]

Confidence interval = 0.5 ± 1.1103

Lower limit = 0.5 - 1.1103 = -0.610

Upper limit = 0.5 + 1.1103 = 1.610

Confidence interval = (-0.610, 1.610)