Question

Pooled‐Variances t‐Test for Mean Comparison We have two normal populations with unknown means μ₁ and μ₂ and a common variance σ². Two independent samples are randomly drawn from the populations with the following data:n₁ = 20 xത₁ = 8.91 s₁ = 2.1n₂ = 24 xത₂ = 8.23 s₂ = 2.5(a)[5] At the level of significance α = 0.01, test H₀: μ₁ = μ₂ versus H₁: μ₁ ≠ μ₂. Sketch the test.

Answer 1

With the p value obtained and using the significance level given
`\alpha=0.01` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 1% of significance the mean of the group 1 is not significantly different from than the mean for the group 2.

Explanation:

1) Notation and hypothesis

When we have two independentt samples from two normal distributions with equal variances we are assuming that

`\sigma^2_1 =\sigma^2_2 =\sigma^2`

And the statistic is given by this formula:

`t=\frac{(\bar X_1 -\bar X_2)-(\mu_(1)-\mu_2)}{S_p\sqrt{(1)/(n_1)}+(1)/(n_2)}`

Where t follows a t distribution with
`n_1+n_2 -2` degrees of freedom and the pooled variance
`S^2_p` is given by this formula:

`\S^2_p =((n_1-1)S^2_1 +(n_2 -1)S^2_2)/(n_1 +n_2 -2)`

This last one is an unbiased estimator of the common variance
`\simga^2`

The system of hypothesis on this case are:

Null hypothesis:
`\mu_1 = \mu_2`

Alternative hypothesis:
`\mu_1 \\eq \mu_2`

Or equivalently:

Null hypothesis:
`\mu_1 - \mu_2 = 0`

Alternative hypothesis:
`\mu_1 -\mu_2 \\eq 0`

Our notation on this case :

`n_1 =20` represent the sample size for group 1

`n_2 =24` represent the sample size for group 2

`\bar X_1 =8.91` represent the sample mean for the group 1

`\bar X_2 =8.23` represent the sample mean for the group 2

`s_1=2.1` represent the sample standard deviation for group 1

`s_2=2.5` represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

`\S^2_p =((20-1)(2.1)^2 +(24 -1)(2.5)^2)/(20 +24 -2)=5.418`

And the deviation would be just the square root of the variance:

`S_p=2.328`

2) Calculate the statistic

And now we can calculate the statistic:

`t=\frac{(8.91-8.23)-(0)}{2.328\sqrt{(1)/(20)}+(1)/(24)}=0.965`

Now we can calculate the degrees of freedom given by:

`df=20+24-2=42`

3) Calculate the p value

And now we can calculate the p value using the altenative hypothesis:

`p_v =2*P(t_(42)>0.965) =0.340`

4) Conclusion

So with the p value obtained and using the significance level given
`\alpha=0.01` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 1% of significance the mean of the group 1 is not significantly different from than the mean for the group 2.