Answer:
98% Confidencce Interval is ( 3030.6, 7467.4 )
Step-by-step explanation:
Given that:
Sample size [tex]n_1 =[/tex] 71
Sample size [tex]n_2 =[/tex] 31
Sample mean [tex]\overline x_1 =[/tex] 41628
Sample mean [tex]x_2 =[/tex] 36,379
Population standard deviation [tex]\sigma_1[/tex] = 4934
Population standard deviation [tex]\sigma_2 =[/tex] 4180
At 98% confidence interval level, the level of significcance = 1 - 0.98 = 0.02
Critical value at [tex]z_{0.02/2} = 2.33[/tex]
The Margin of Error = [tex]z \times \sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma^2_2}{n_2} }[/tex]
= [tex]2.33 \times \sqrt{\dfrac{4934^2}{71}+\dfrac{4180^2}{31} }[/tex]
= [tex]2.33 \times \sqrt{\dfrac{24344356}{71}+\dfrac{17472400}{31} }[/tex]
= [tex]2.33 \times \sqrt{906504.06 }[/tex]
= 2218.40
The Lower limit = [tex]( \overline x_1 - \overline x_2) - (Margin \ of \ error)[/tex]
= ( 41628 - 36379 ) - ( 2218.40)
= 5249 - 2218.40
= 3030.6
The upper limit = [tex]( \overline x_1 - \overline x_2) + (Margin \ of \ error)[/tex]
= ( 41628 - 36379 ) + ( 2218.40)
= 5249 + 2218.40
= 7467.4
∴ 98% Confidencce Interval is ( 3030.6, 7467.4 )
