Question

Population 1 has an estimated mean of 50 and a sample standard deviation of 3 while population 2 has a sample mean of 48 and a sample standard deviation of 4. A sample of size 16 is taken for each population.

Given that the sample sizes are small, we have to "pool the variances," operating on the assumption that the variances are equal. What is the value of pooled variance?

What is the standard error of the difference in means, µ1 - µ2?

The "critical" absolute value of "t" for the null hypothesis that the means are equal versus the alternative of not equal for "n1+n2 -2" degrees of freedom at the 95% level of confidence would be?

The "calculated" value of "t" for this situation would be?

Answer 1

Given that the sample sizes are small, we have to "pool the variances," operating on the assumption that the variances are equal. What is the value of pooled variance?

Formula for pooled variance is given as below:

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

We are given

X1bar = 50

X2bar = 48

S1 = 3

S2 = 4

n1 = 16

n2 = 16

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

S_p² = [(16 – 1)*3^2 + (16 – 1)*4^2]/(16 + 16 – 2)

S_p² = 12.5000

Pooled variance = 12.5

What is the standard error of the difference in means, µ1 - µ2?

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

Standard error = sqrt[12.5*((1/16)+(1/16))]

Standard error = 1.25

The "critical" absolute value of "t" for the null hypothesis that the means are equal versus the alternative of not equal for "n1+n2 -2" degrees of freedom at the 95% level of confidence would be?

df = n1 + n2 – 2 = 16 + 16 – 2 = 30

Confidence level = 95% = 0.95

Level of significance = α = 1 – 0.95 = 0.05

Critical absolute t value = 2.0423

(by using t-table)

The "calculated" value of "t" for this situation would be?

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

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

t = (50 – 48) / sqrt[12.5*((1/16)+(1/16))]

t = 2/1.25

t = 1.60

Calculated value of t = 1.60

P-value = 0.1201

P-value > α = 0.05

So, we do not reject the null hypothesis