Since is a right tailed test we need to look on the t distribution with 12 degrees of freedom that accumulates 0.05 of the area on the right. And we can use the following excel code:

"=T.INV(0.95,12)" and we got $t_{crit}=+1.782$

Step-by-step explanation:

When we have two independent 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 $\sigma^2$

The system of hypothesis on this case are:

Null hypothesis: $\mu_O \leq \mu_E$

Alternative hypothesis: $\mu_O > \mu_E$

Or equivalently:

Null hypothesis: $\mu_O - \mu_E \leq 0$

Alternative hypothesis: $\mu_O -\mu_E > 0$

Our notation on this case :

$n_E =6$ represent the sample size for group Edne

$n_O =8$ represent the sample size for group Orno

$\bar X_E =7$ represent the sample mean for the group Edne

$\bar X_O =10$ represent the sample mean for the group Orno

$s_E=1.414$ represent the sample standard deviation for group Edne

$s_O=2.268$ represent the sample standard deviation for group Orno

First we can begin finding the pooled variance:

$\S^2_p =\frac{(6-1)(1.414)^2 +(8 -1)(2.268)^2}{6 +8 -2}=3.834$

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

$S_p=1.958$

And now we can calculate the statistic:

$t=\frac{(10 -7)-(0)}{1.958\sqrt{\frac{1}{6}+\frac{1}{8}}}=2.837$

Now we can calculate the degrees of freedom given by:

$df=6+8-2=12$

Since is a right tailed test we need to look on the t distribution with 12 degrees of freedom that accumulates 0.05 of the area on the right. And we can use the following excel code:

"=T.INV(0.95,12)" and we got $t_{crit}=+1.782$

Since our calculated value is higher than the critical value we have anough evidence to reject the null hypothesis, and on thsi case the mean for Orno is significanlty higher than the mean for Edne at 5% of significance.

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3 years ago