A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.69 hours, with a standard deviation of 2.42 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.32 hours, with a standard deviation of 1.83 hours.
Construct and interpret a 95% confidence interval for the mean difference in leisure time between adults with no children and adults with children (μ1 - μ2).

Following are the calculation to the confidence interval:

Given:

To find:

confidence interval=?

Solution:

           (from standard normal table)

Calculating the confidence interval when its value is :

Therefore, the final answer is "(2.159 and 0.581)".

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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

95% of confidence intervals for (μ₁ - μ₂)are determined by

where

The standard error of the difference between two means

Degrees of freedom γ = n₁ +n₂ -2 = 40+40 -2 =78

t₀.₀₂₅ = 1.992

95% of confidence intervals for (μ₁ - μ₂)are determined by

(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)

Conclusion:-

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)

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step-by-step explanation: