Question

Assume that you have a sample of n 1 equals 8n1=8​, with the sample mean Upper X overbar 1 equals 42X1=42​, and a sample standard deviation of Upper S 1 equals 4S1=4​, and you have an independent sample of n 2 equals 15n2=15 from another population with a sample mean of Upper X overbar 2 equals 34X2=34 and a sample standard deviation of Upper S 2 equals 5S2=5. What assumptions about the two populations are necessary in order to perform the​pooled-variance t test for the hypothesis Upper H 0 : μ 1 equals μ 2H0: μ1=μ2 against the alternative Upper H 1 : μ 1 >μ 2H1: μ1>μ2 and make a statistical​ decision?

Answer 1

Check the explanation below

Explanation:

Hello!

To make a pooled variance t-test you have to make the following assumptions:

The study variables X₁ and X₂ must be independent.

Both variables should have a normal distribution, X₁~N(μ₁; σ₁²) and X₂~N(μ₂; σ₂²)

The population variances should be equal but unknown, σ₁² = σ₂² = ?.

You have the information of two samples:

Sample 1

n₁=8

sample mean X[bar]₁= 42

sample standard deviation S₁=4

Sample 2

n₂=15

sample mean X[bar]₂= 34

sample standard deviation S₂= 5

For the hypothesis:

H₀: μ₁ = μ₂

H₁: μ₁ > μ₂

The statistic is:

t= (X[bar]₁ - X[bar]₂) - (μ₁ - μ₂) ~
`t_(n_1 + n_2 - 2)`

Sa
`\sqrt{(1)/(n_1) + (1)/(n_2) }`

Sa²=
`((n_1-1)S_1^2+(n_2-1)S_2^2)/(n_1+n_2-2)`

Sa²= 22

Sa= 4.69

`t_(H0)`= 3.8962 ≅ 3.9

The critical region is one-tailed, for example for α: 0.05

`t_(n_1 + n_2 - 2; 1 - \alpha ) = t_(21; 0.95) = 1.721`

Since
`t_(H0)` > 1.721, then the decision is to reject the null hypothesis.

I hope it helps!