Answer:
95% confidence interval for the mean difference in leisure time between adults with no children and adults with children (μ1 - μ2).
(0.4144 , 2.3256)
Step-by-step explanation:
Given sample size 'n' =n₁ = n₂ = 40
The mean of the first sample (x₁⁻) = 5.69 hours
The standard deviation of the first sample (S₁)= 2.42 hours
The mean of the second sample( x₂⁻) = 4.32 hours
The standard deviation of the second sample (S₂)= 1.83 hours
<em>95% of confidence intervals for (μ₁ - μ₂)are determined by</em>
<em></em><em></em>
<em>where </em>
<em>The standard error of the difference between two means </em>
<em></em><em></em>
<em></em><em></em>
<em></em><em></em>
<em>Degrees of freedom γ = n₁ +n₂ -2 = 40+40 -2 =78</em>
t₀.₀₂₅ = 1.992
<em>95% of confidence intervals for (μ₁ - μ₂)are determined by</em>
<em></em><em></em>
(5.69 -4.32)- 1.992(0.47972)), (5.69-4.32)+1.992(0.47972))
(1.37 -0.9556 , 1.37+0.9556)
(0.4144 , 2.3256)
<u><em>Conclusion</em></u>:-
95% confidence interval for the mean difference in leisure time between adults with no children and adults with children (μ1 - μ2).
(0.4144 , 2.3256)