10 Two separate samples receive different treatments After t

10. Two separate samples receive different treatments. After treatment, the first sample has n = 9 with SS = 462 , and the second has n = 7 with SS = 420 . Compute the pooled variance for the two samples. Calculate the estimated standard error for the sample mean difference. If the sample mean difference is 10 points, is this enough to reject the null hypothesis using a two-tailed test with a = .05 ?

Solution

We have given that two separate samples receive different treatments.

After treatment, the first sample has n1 = 9 with SS = 462

and the second has n2 = 7 with SS = 420

We have to calculate pooled variance for the two samples.

For the first sample n = 9 with SS = 462

The sample variance can be calculated by using the formula,

S1² = SS / n1-1 = 462 / 9-1 = 462 / 8 = 57.75

and for the second sample n = 7 with SS = 420

S2² = SS / n2 - 1 = 420 / 7 - 1 = 420 / 6 = 70

The pooled variance is,

Sp² = [ (n1 - 1)*S1² + (n2 - 1)*S2² ] / n1 + n2 - 2

= [ ( 9 - 1) * 57.75 + (7 - 1) * 70 ] / 9 + 7 - 2

= (8 * 57.75 + 6 * 70) / 14

= 63

This is the pooled variance.

Calculate the estimated standard error for the sample mean difference.

Now pooled standard deviation (Sp) = sqrt(pooled variance)

Pooled standard deviation (Sp) = sqrt( 63) = 7.9373

Pooled standard error = Sp * sqrt(1/n1 + 1/n2)

= 7.9373 * sqrt( 1 / 9 + 1 / 7 )

= 7.9373 * 0.50395 = 4.00000

If the sample mean difference is 10 points,

t = sample mean difference / pooled standard error

= 10 / 4.00000 = 2.5

Is this enough to reject the null hypothesis using a two-tailed test with a = .05 ?

That means we have given that the hypothesis is two sided,

H₀ : µ1 = µ2 Vs H₁ : µ1 µ2

a = 0.05

We can find P-value by using EXCEL.

syntax : tdist(x , d.f. , tails)

x = 2.5

d.f. = 14

tails = 2

This will gives us P-value = 0.025

P-value < a

Reject the null hypothesis at 5% level of significance.