Question

Assume that you have a sample size of n₁ = 16 with a mean of 42 and a standard deviation (S) equal to 9. Assume that you have another independent sample with n₂ = 25, a mean of 36 and a standard deviation (S) of 4. Assume you are directed to use a significance level of α = 0.01. [DM.4]

1. Construct the appropriate hypothesis test.  Identify H0 and H1.
2. What are the appropriate critical values? (4 Decimal Places)
3. From what distribution (z or t)? If t, with how many degrees of freedom?
4. What is the difference between means? (2 Decimal Places)
5. What is the standard error of the difference between the means? (3 Decimal Places)
6. What is the appropriate test statistic? (4 Decimal Places)
7. What is your statistical decision?

Answer 1

Construct the appropriate hypothesis test.  Identify H0 and H1.

Here, we have to use two sample t test for population means.

H₀: µ₁ = µ₂ versus H₁: µ₁ ≠ µ₂

What are the appropriate critical values? (4 Decimal Places)

We are given n1 = 16, n2 = 25

df = n1 + n2 – 2 = 16 + 25 – 2 = 39

α = 0.01

Critical values = -2.7079 and 2.7079

(by using t-table)

From what distribution (z or t)? If t, with how many degrees of freedom?

t distribution with df = 39

What is the difference between means? (2 Decimal Places)

We are given X1bar = 42, X2bar = 36

Difference between means = X1bar – X2bar = 42 – 36 = 6

What is the standard error of the difference between the means? (3 Decimal Places)

Standard error = 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 S1 = 9, S2 = 4, n1 = 16, n2 = 25

S_p² = [(16 – 1)*9^2 + (25 – 1)*4^2]/(16 + 25 – 2)

S_p² = 41

Standard error = sqrt[S_p²*((1/n1)+(1/n2))]

Standard error = sqrt[41*((1/16)+(1/25))]

Standard error = 2.0500

What is the appropriate test statistic? (4 Decimal Places)

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

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

t = 6/2.0500

t = 2.9268

What is your statistical decision?

From above parts, we have

P-value = 0.0057

(by using t-table)

P-value < α = 0.01

So, we reject the null hypothesis

There is sufficient evidence to conclude that both population means are different.