Question

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).

Answer 1

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

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

$(X^{-} _{1} - X^{-} _{2} - t_{\frac{\alpha }{2} } Se(X^{-} _{1} - X^{-} _{2} ) , X^{-} _{1} - X^{-} _{2} + t_{\frac{\alpha }{2} } se(X^{-} _{1} - X^{-} _{2} ))$

where

The standard error of the difference between two means

$se(X^{-} _{1} - X^{-} _{2} ) = \sqrt{\frac{S^{2} _{1} }{n_{1} }+\frac{S^{2} _{2} }{n_{2} } }$

$se(X^{-} _{1} - X^{-} _{2} ) = \sqrt{\frac{(2.42)^2 }{40 }+\frac{(1.83)^2 }{40 } }$

$se(X^{-} _{1} - X^{-} _{2} ) = \sqrt{0.2301325} = 0.47972$

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

$t_{\frac{\alpha }{2} } = t_{\frac{0.05}{2} } = t_{0.025}$

t₀.₀₂₅ = 1.992

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

$(X^{-} _{1} - X^{-} _{2} - t_{\frac{\alpha }{2} } Se(X^{-} _{1} - X^{-} _{2} ) , X^{-} _{1} - X^{-} _{2} + t_{\frac{\alpha }{2} } se(X^{-} _{1} - X^{-} _{2} ))$

(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)

Answer 2

Following are the calculation to the confidence interval:

Given:

$\bar{x_1}= 5.69\\\\\bar{x_2}= 4.32\\\\s_1=2.42\\\\s_2=1.83\\\\n_1=40\\\\n_2=40$

To find:

confidence interval=?

Solution:

$\to a=0.1\\\\ \to Z(0.05)=1.645$            (from standard normal table)

Calculating the confidence interval when its value is $95\%$:

$\to (\bar{x_1}-\bar{x_2}) \pm Z \times \sqrt{(\frac{s^2_{1}}{n_1}+ \frac{s^2_{2}}{n_2})}$  

$\to (5.69-4.32)\pm 1.645 \times \sqrt{(\frac{2.42^2}{40}+\frac{1.83^2}{40})}\\\\\to (1.37)\pm 1.645 \times \sqrt{(\frac{5.8564}{40}+\frac{3.3489}{40})}\\\\\to (1.37)\pm 1.645 \times \sqrt{(\frac{5.8564+3.3489}{40})}\\\\\to (1.37)\pm 1.645 \times \sqrt{(\frac{9.2053}{40})}\\\\\to (1.37)\pm 1.645 \times \sqrt{0.2301325}\\\\\to (1.37)\pm 1.645 \times 0.4797212 \\\\\to (1.37)\pm 0.789\\\\\to (2.159, 0.581 )$

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

Learn more about the confidence interval:

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