Question

The null and alternate hypotheses are:

H₀ : μ₁ = μ₂
H₁ : μ₁ ≠ μ₂

A random sample of 9 observations from one population revealed a sample mean of 24 and a sample standard deviation of 3.7. A random sample of 6 observations from another population revealed a sample mean of 28 and a sample standard deviation of 4.6.

At the 0.01 significance level, is there a difference between the population means?

a. State the decision rule.

b.Compute the pooled estimate of the population variance.

c. Compute the test statistic.

d. State your decision about the null hypothesis.

e. The p-value is

Answer 1

Solution:
Null hypothesis is H0: mean 1 = mean 2
Alternate hypothesis H1: mean1 is not equal to mean 2
Here degree of freedom = (6+6-2) = 10
this is two tailed test and alpha=0.01
so tcritical value is +/-3.1692
We will reject null hypothesis if test statistic is greater than 3.1692 or less than -3.1692
Pooled estimate varinace can be estimated as
Sp^2 = ((n1-1)S1^2 +(n2-1)S2^2)/(n1-1)+(n2-1) = ((6-1)*3.7*3.7 +(6-1)*4.6*4.6)/((6-1)+(6-1) = 174.25/10 = 17.425

test statistic = (X1bar - X2bar) - (mean1-Mean2) / Sqrt(Sp^2 *((1/n1)+(1/n2))
= (24-28)/sqrt(17.425*(1/6 +(1/6)
= -4/2.41 = -1.66

Here we can see that the test statistic is in between +/- 3.1692 so we are failed to reject the null hypothesis, and this result is not significnat, we dont have enough evidence for alternate hypothesis.

p-value is 0.1279