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

Answer:

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.

Step-by-step 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:

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

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:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

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: $\mu_1 \neq \mu_2$

Or equivalently:

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

Alternative hypothesis: $\mu_1 -\mu_2 \neq 0$

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:

$\S^2_p =\frac{(20-1)(2.1)^2 +(24 -1)(2.5)^2}{20 +24 -2}=5.418$

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:

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

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.