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 5.35 hours, with a standard deviation of 2.39 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.47 hours, with a standard deviation of 1.78 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with children left parenthesis mu 1 minus mu 2 right parenthesis12. Let mu 11 represent the mean leisure hours of adults with no children under the age of 18 and mu 22 represent the mean leisure hours of adults with children under the age of 18. The 90% confidence interval for left parenthesis mu 1 minus mu 2 right parenthesis12 is the range from nothing hours to nothing hours. (Round to two decimal places as needed.) What is the interpretation of this confidence interval? A. There is 90% 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 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. C. There is 90% confidence that the difference of the means is in the interval. Conclude that there is aa significant difference in the number of leisure hours. D. 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.

Explanation / Answer

From followiing output we have 90% Confidence interval =(0.096, 1.664)

So, C. There is 90% confidence that the difference of the means is in the interval. Conclude that there is aa significant difference in the number of leisure hours.

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean
1 40 5.35 2.39 0.38
2 40 4.47 1.78 0.28

Difference = (1) - (2)
Estimate for difference: 0.880
90% CI for difference: (0.096, 1.664)
T-Test of difference = 0 (vs ): T-Value = 1.87 P-Value = 0.066 DF = 78
Both use Pooled StDev = 2.1072

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