Question

A random sample of 40 adults with no children under the age of 18 years results in a meandaly leisure time of 5.92 hours, with a standard deviation of 2.33 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.26 hours, with a standard deviation of 1.61 hours. Construct and interpret a 95% confidence interval for the mean difference in leisure time between adults with no children and adults with children (14-12) Let Hy represent the mean leisure hours of adults with no children under the age of 18 and 12 represent the mean leisure hours of adults with children under the age of 18. The 95%. cfidence interval for (4 - 2) is the range from hours to hours. (Round to two decimal places as needed.) What is the interpretation of this confidence interval? O A There is sx confidence 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. OB. There is 96% confidence that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours OC. There is a 95% probability that the difference of the means is in the interval Conclude that there is insufficient evidence of a significant Gifference in the number of our hours. D. There is a 6% probability that the difference of the means is in the interval. Conclude that there is a significant atterence in the number Click 10 let your answer hilar MacBook Pro 3 $ 4 un 6 & 7 8 Q W E R T U caps lock А S D F G H N shift Х. C V B Ν M A control option command

Answer 1

Sample #1   ---->   1
mean of sample 1,    x̅1=   5.92
standard deviation of sample 1,   s1 =    2.3300
size of sample 1,    n1=   40
      
Sample #2   ---->   2
mean of sample 2,    x̅2=   4.260
standard deviation of sample 2,   s2 =    1.61
size of sample 2,    n2=   40

Degree of freedom, DF=       69
t-critical value = t α/2 =    1.995   (excel formula =t.inv(α/2,df)
      
std error , SE =    √(s1²/n1+s2²/n2) =    0.448
margin of error, E = t*SE =    1.995*0.4478=   0.893337
      
difference of means = x̅1-x̅2 =    5.92-4.26=   1.6600
confidence interval is       
Interval Lower Limit = (x̅1-x̅2) - E =    1.66-0.8933=   0.77
Interval Upper Limit = (x̅1-x̅2) + E =    1.66+0.8933=   2.55

(0.77 , 2.55)

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

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