Question

Was the A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.96 hours, with a standard deviation of 226 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4 23 hours, with a standard deviation of 155 hours. Construct and interpret a 95% confidence interval for the mean difference in leisure time between adults with no children and adults with children (1) Let Hy represent the mean loisure 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 ion V ut this hours rento The 95% confidence interval for (-; --2) is the range from hours to (Round to two decimal places as needed) Lion

Answer 1

we need to couetruet ورو) 95% confidence interval fero difference between that Population mean (41-42) for tere not known Population standart daviation given 5.96 X2 423 S2 1.55 si = 2.26 HO a uz 40 ale ume we tee information provided Based so there aree equal, variance that the population free dow dfa 40 tho-2 tere degree of 78 ts critical for 2 df xao.05 The value the Сс Iggl . tere since Population variance aree assumed tere pooled to to coupute we equal need standard deviation (n.-1) sirt (Mal) să Sp nit 2-2 (40-1) 2.26²f (40-1) 1.552 40+ 40- = 1938

since ee assumed that Population standard error radiance tre equal 2 as follows is combuted th SP se p. 938 HO 35/76 + = 0.433 X, X2 + tex sed c I 1) (x,-X2 - texse 1 (5.96-4:23-1.981x 0 433 5:96-4 23+1'991*0'433) 2 (0.87, 2.59) The 954. Confidence internal tor (41-4₂) is thee from 0.87 hours to range 2-59 hours