A random sample of size $n_1 = 25,$ taken from a normal population with a standard deviation $\sigma_{1}=5.2$ has a mean $\overline{x}_{1}=81.$ A second random sample of size $n_{2}=36,$ taken from a different normal population with a standard deviation $\sigma_{2}=3.4,$ has a mean $\overline{x}_{2}=76$ Test the hypothesis that $\mu_{1}=\mu_{2}$ against the alternative, $\mu_{1} \neq \mu_{2}.$ Quote a P-value in your conclusion.

Solution

VerifiedWe have:

$n_{1}=25, \\ \sigma_{1}=5.2, \\ \overline{x}_{1}=81$

$n_{2}=36, \\ \sigma_{2}=3.4, \\ \overline{x}_{2}=76$

The null hypothesis is:

$H_{0} : \mu_{1}=\mu_{2}$

The alternative hypothesis is:

$H_{1} : \mu_{1} \neq \mu_{2}$

Therefore, we have

$H_{0} : \mu_{1}-\mu_{2}=0$

And

$H_{1} : \mu_{1}-\mu_{2} \neq 0$

We will use the level of significance $\alpha=0.05$