Question

Two separate samples receive different treatments. After treatment, the first sample has n = 6 with SS = 236, and the second has n 5=12 with SS = 340.

a. Find the pooled variance for the two samples.
b. Compute the estimated standard error for the sample mean difference.

Answer 1

The pooled variance for the two samples is 320.25. The estimated standard error for the sample mean difference is approximately 8.228.

Step-by-step explanation:

To find the pooled variance for the two samples, we can use the formula:

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

Substituting the given values, we have:

Sp² = [((6-1) * 236) + ((12-1) * 340)] / (6 + 12 - 2)

Sp² = 1176 + 3948 / 16

Sp² = 5124 / 16

Sp² = 320.25

Therefore, the pooled variance for the two samples is 320.25.

To compute the estimated standard error for the sample mean difference, we can use the formula:

SE = sqrt[(S1² / n1) + (S2² / n2)]

Substituting the given values, we have:

SE = sqrt[(236 / 6) + (340 / 12)]

SE = sqrt[39.33 + 28.33]

SE = sqrt[67.66]

SE ≈ 8.228

Therefore, the estimated standard error for the sample mean difference is approximately 8.228.

Answer 2

Answer: a) 33.88 and b) 2.91.

Step-by-step explanation:

Since we have given that

Sample n₁ = 6

SS = 236

Sample n₂ = 12

SS = 340

So, we need to find the following:

a. Find the pooled variance for the two samples.

`S_(p^2)=(SS_1+SS_2)/(n_1+n_2-1)`

So, it becomes,

`S_p^2=(340+236)/(12+6-1)=(576)/(17)=33.88`

b. Compute the estimated standard error for the sample mean difference.

`SE=S_p* \sqrt{(1)/(n_1)+(1)/(n_2)}\\\\SE=√(33.88)* \sqrt{(1)/(6)+(1)/(12)}\\\\SE=5.82* \sqrt{(2+1)/(12)}\\\\SE=5.82* \sqrt{(3)/(12)}\\\\SE=5.82* \sqrt{(1)/(4)}\\\\SE=5.82* (1)/(2)\\\\SE=2.91`

Hence, a) 33.88 and b) 2.91.