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This Question: 1 pt A random sample of 40 adults with no children under the age of 18 years results in a mandaly sure time of
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Answer #1

Adults with no children,

number of samples, n= 40

Mean, \mu_1 = 5.55 \; hours

standard\; deviation,\sigma =2.38 \;hours

Adults with children

number of samples, n= 40

Mean, \mu_2 = 4.32 \; hours

standard\; deviation,\sigma =1.53 \;hours

The 95% confidence interval for (\mu_1 - \mu_2) is given by,

Confidence\;interval = (\mu_1-\mu_2)\pm z*\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}

Confidence\;interval = (5.55-4.32)\pm 1.96*\sqrt{\frac{2.38^2}{40}+\frac{1.53^2}{40}}

Confidence\;interval = [0.35, 2.12]

Therefore,

The 95% confidence interval for (\mu_1 - \mu_2) is the range from 0.35 hours to 2.21 hours

The interpretation is,

D. There is 95% confidence that the diffrence of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours.

Reason : Since the confidence interval does not include 0 , we can conclude that there is significant difference in the means.

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