Assuming that we are dealing with a normal distribution.
I have two different groups and I want to compare the means between this two independent groups.
Group 1 has $ \bar X_1 =2.60$, and I have calculated the Standard deviation from that sample which is $\sigma_1= 0.56$;
Group 2 has $ \bar X_2 =1.30 $, I have calculated the Standard deviation from the sample $\sigma_2= 1.02$;
The sample size for both are $n=12$.
a) Using parametric tests I need to know whether the mean is lower in the second group than in the first group. By $t$-distribution tables I need to know the approximate $p$-value for this.
b) Find $95\%$ CI for the difference in the mean the two groups.
My work:
a) I am not sure how to do that but I think I can use this formula:
\begin{align*} z&= \frac{\mu_1-\mu_2}{\sqrt{\sigma_1^2/n_1+\sigma_2^2/n_2}}\\ &= \frac{2.60-1.30}{\sqrt{\frac{0.56^2}{12}+\frac{1.02^2}{12}}}\\ &=3.9 \end{align*} and then $$P(z>3.9)=1-0.999952=0.0000048,$$ so the $p$-value is $0.0000048$. Because $p <0.001$ there is a strong difference between the two groups.
But I am not really sure.
b) I did:
formula: $$\mu_1-\mu_2 \pm(1.96)\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$$
$ \mu_1-\mu_2=2.60-1.30=1.3$
$95\%$CI=$1.3 \pm(1.96)\sqrt{\frac{0.56^2+1.02^2}{12}} =$
$ 1.3\pm (0.65)\to$ that gives the interval $(0.65 ,1.96)$.
So, the confidence interval is
$$95\% \text{ CI}=(0.65 ,1.96)$$
Can anyone let me know if I am doing it correctly?