Answer:
The 95% confidence interval for the difference in population means is (−26.325175 , 36.325175)
Step-by-step explanation:
Given that :
sample size n₁ = 36
sample mean [tex]\over\ x[/tex]₁ = 83
standard deviation [tex]\sigma[/tex]₁ = 5
sample size n₂ = 49
sample mean [tex]\over\ x[/tex]₂= 78
standard deviation [tex]\sigma[/tex]₂ = 3
The objective is to construct a 95% confidence interval for the difference in the population means
Let the population means be [tex]\mu_1[/tex] and [tex]\mu_2[/tex]
The 95% confidence interval or the difference in population means can be calculated by using the formula;
[tex](\overline{x_1} - \overline{x_2}) \pm t_{\alpha /2} \ \times s_{p}[/tex]
where;
the pooled standard deviation [tex]s_{p} = \dfrac{(n_1-1)s_1^2+(n_2-1)s^2_2}{n_1+n_2-2}[/tex]
[tex]s_{p} = \dfrac{(36-1)5^2+(49-1)3^2}{36+49-2}[/tex]
[tex]s_{p} = \dfrac{(35)25+(48)9}{83}[/tex]
[tex]s_{p} = \dfrac{875+432}{83}[/tex]
[tex]s_{p} = \dfrac{1307}{83}[/tex]
[tex]s_p[/tex] = 15.75
degree of freedom = [tex]n_1 +n_2 -2[/tex]
degree of freedom = 36+49 -2
degree of freedom = 85 - 2
degree of freedom = 83
The Critical t- value 95% CI at df = 83 is
t critical = T.INV.2T(0.05, 83) = 1.9889
Therefore, for the population mean , we have:
= (83 - 78) ± (1.9889 × 15.75)
= 5 ± 31.325175
= 5 - 31.325175 , 5 + 31.325175
= (−26.325175 , 36.325175)
