A random sample of 40 adults with no children under the age of 18 years results

Question

A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.19 hours, with a standard deviation of 2.49 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.18 hours, with a standard deviation of 1.75 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with children (mu 1-mu 2). Let mu 1 represent the mean leisure hours of adults with no children under the age of 18 and mu 2 represent the mean leisure hours of adults with children under the age of 18. The 90% confidence interval for (mu 1-mu 2) is the range from ___ to ___ hours (round to 2 decimal places as needed) What is the interpretation of this confidence interval? There is a 90% probability that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours. There is 90% confidence that the difference of the means is in the interval. Conclude that there is insufficient evidence of a insufficient evidence of a significant difference in the number of leisure hours. There is a 90% probability that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours. There is 90% confidence that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours.

Explanation / Answer

Answer

Given,

Sample 1

Sample size(n1)=40

Mean(x1)=5.19

Standard Deviation(SD1)=2.49

Sample 2

Sample size(n2)=40

Mean(x2)=4.18

Standard Deviation(SD2)=1.75

Confidence Interval is 90%

Hence corresponding Z-value is 1.645

The formula for constructing confidence interval is,

[x1-x2] +/- Z*[sqrt{SD1*SD1/n1+SD2*SD2/n2}]

X1-x2=5.19-4.18=1.01

SD1*SD1/n1=2.49*2.49/40=0.155

SD2*SD2/n2=1.75*1.75/40=0.0766

SD1*SD1/n1+SD2*SD2/n2=0.155+0.0766=0.2315

sqrt{SD1*SD1/n1+SD2*SD2/n2=0.48

Z* sqrt{SD1*SD1/n1+SD2*SD2/n2=0.48*1.645=0.7916

Hence confidence interval is

1.01-0.7916 µ1-µ2 1.01+0.7916

Confidence Interval is 0.2184 µ1-µ2 1.8016

We can signify with 90% confidence that the difference of mean hours is between 0.2184 and 1.8016 interval

Also, both the upper limit and lower limit of the confidence Interval are positive

Hence there is a significant difference in the leisure hours

Therefore, the answer would be D

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