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

5.83

hours, with a standard deviation of

2.44

hours. A random sample of

40

adults with children under the age of 18 results in a mean daily leisure time of

4.42

hours, with a standard deviation of

1.56

hours. Construct and interpret a

95%

confidence interval for the mean difference in leisure time between adults with no children and adults with children

(12)

Let

1

represent the mean leisure hours of adults with no children under the age of 18 and

2

represent the mean leisure hours of adults with children under the age of 18.The

95%

confidence interval for

(12)

is the range from

hours to

hours.

(Round to two decimal places as needed.)

What is the interpretation of this confidence interval?

A.

There is

95%

confidence that the difference of the means is in the interval. Conclude that there is

insufficient evidence of ainsufficient evidence of a

significant difference in the number of leisure hours.

B.

There is

95%

confidence that the difference of the means is in the interval. Conclude that there is

aa

significant difference in the number of leisure hours.

C.

There is a

95%

probability that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours.

D.

There is a

95%

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.

Explanation / Answer

Calculate Confidence Interval

t/2 = 1.99081

Standard Error = [(n-1)(s)² + (n-1)(s)²]/[n + n - 2] • 1/n + 1/n
= 4.1936 0.05
= 0.4579

Lower Bound = (x - x) - t/2•([(n-1)(s)² + (n-1)(s)²]/[n + n - 2]•1/n + 1/n)
= (5.83 - 4.42) - (1.99081)(0.45790)
= 0.4983

Upper Bound = (x + x) + t/2•([(n-1)(s)² + (n-1)(s)²]/[n + n - 2]•1/n + 1/n)
= (5.83 - 4.42) + (1.9908)(0.4579)
= 2.3216

Confidence Interval = (0.4983, 2.3216)



Interpretation of a confidence interval:
Since we do not know if the confidence interval (0.4983, 2.3216) contains ( - ) or not, we are only 95% confident that (0.4983, 2.3216) contains ( - ).

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