Question

A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.69 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 4.07 hours, with a standard deviation of 1.54 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with children (H1-H2) Let y 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 (H1-H2) is the range from hours to hours, (Round to two decimal places as needed.)

Answer 1

Answelt Given that nia no 77 519.5 = 2149. 2 = uo, n=1107 S2 = 1.54. a= -8,90 = 0110 d t = nit n-3 = 78 teledt = 3.9 qoy confidence interval tol population mean. (Mi-12) in (m1-2) I tapet & spr cyn it you) where spr= (h.-) s, if Cn2 = 1) {" nit n2-2. = 39 C 2.099 +(39) (115u5r 78. - u 1285 [ribe) 239 Euro ( 10 ) = +621 34 7 ON ] -) 162+178. (01 1672,-0116) 90% contidence is teeval tol y la Z 0. 1672 hours to 0.16 hour. The