Is there a significant difference between American men and American women in their attitude towards President Donald Trump? You asked a random sample of 77 American voters to rate their level of support for Donald Trump on a scale of 1-100 (with higher scores representing a higher level of support). The 38 men in the sample had a mean score of 70.3 with a standard deviation of 5.3. The 39 women in the sample had a mean score of 68.9 with a standard deviation of 5.7. Is there a significant difference between men and women on this variable?
If anyone can help me answer this question it would be gratefully appreciated. I have been trying to do practice hypothesis testing questions but I do not understand how to get the t critical. I understand how to extract the information from the question and then how to solve for the degree of freedom [n1+n2-2], but I do not understand how the t critical is derived from that. I also do not understand what alpha would be. Someone please explain this.
Let $$H_0 : \mu_1 = \mu_2$$
$$H_a : \mu_1 \neq \mu_2$$
We have $\bar{X}_1=70.3$, $s_1=5.3$, $n_1=38$ and $\bar{X}_2=68.9$, $s_2=5.7$, $n_2=39$
With the assumption of equal population variances we have
$$\begin{align*} s_{pooled}^2 &=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}\\\\ &=\frac{(38-1)\cdot5.3^2+(39-1)\cdot5.7^2}{38+39-2}\\\\ &\approx 30.32 \end{align*}$$
Then
$$\frac{\bar{X}_1-\bar{X}_2}{s_{pooled}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\sim t_{n_1+n_2-2}$$
and so we have
$$\frac{70.3-68.9}{\sqrt{30.32}\sqrt{\frac{1}{38}+\frac{1}{39}}}\approx1.115\sim t_{75}$$
Since this is a two-sided test, we compare this test statistic with the critical value at $$\alpha=\frac{0.05}{2}=0.025$$
By a t-table we get
$$t_{75,0.025}\approx 1.96$$
Since $1.115\lt1.96$ we fail to reject the null hypothesis at $\alpha=0.05$
Alternatively, software gives that a t-value of $1.115$ with $df=75$ gives a two-sided p-value of $0.268$.
Either way, we do not have significant evidence that the population means differ.