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 5.05 hours, with a standard deviation of 2.42 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.22 hours, with a standard deviation of 1.87 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with children (1 2). Let 1 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 90% confidence interval for (12) is the range from _______ to _______ . Please solve using technology and show the steps.

Explanation / Answer

Using minitab,

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean
1 40 5.05 2.42 0.38
2 40 4.22 1.87 0.30

Difference = (1) - (2)
Estimate for difference: 0.830
90% CI for difference: (0.024, 1.636)
T-Test of difference = 0 (vs ): T-Value = 1.72 P-Value = 0.090 DF = 73

The 90% confidence interval for (12) is the range from 0.024 to 1.636 .

Steps :

To calculate a confidence interval or perform a hypothesis test for the difference in the means from two independent populations:

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