Question

Is there an association between napping and “inactive” exercise by testing the difference of proportions using a z-score. By following your steps for hypothesis testing, conduct a test to determine if there is a significant difference between inactive exercise rates and napping.

Age Table 1 Baseline characteristics of the participants according to daytime napping Daytime napping, no. (%) p-Value No Yes 9613 (71.4) 3852 (286) 65 6794 (70.7) 1631 (423) <0.001 >65 2819 (293) 2221 (57.7) Sex Men 3789 (394) 2098 (545) 0.001 Women 5824 (60.6) 1754 (45.5) Education Lower 4019 (418) 1819 (472) 0.001 Higher 5594 (582) 2033 (528) Marital status Single 375 (39) 161 (4.2) Married 7795 (81.1) 3070 (79.7) Others 1443 (150) 621 (16.1) Smoking status Current smokers 764 (79) 329 (8.5) <0.001 Former smokers 3841 (40.0) 1873 (486) Never smoked 5008 (52.1) 1650 (428) Alcohol intake 35 4850 (505) 1975 (51.3) >3.5 4763 (495) 1877 (48.7) Physical activity Inactive 2287 (238) 1223 (31.7) 0001 Moderately inactive 2938 (30.6) 1052 (273) Moderately active 2473 (25.7) 846 (220) Active 1915 (199) 731 (19.0) Hypnotic drug use No 9458 (98.4) 3762 (977) 2001 155 (16) 90 (23) Pre-existing diseases 7188 (748) 2656 (69.0) <0.001 Yes 2425 (252) 1196 (31.0) Widowed, separated or divorced. Units/week, divided by median Yes NO

Answer 1

Date: 11/11/2019 Answer To test the hypothesis is that there is a significant difference between inactive exercise rates and napping at 5% significance level The null and alternative hypothesis is, Ho P P2 . a The z-test statistics is By using Excel software, find z-test statistics with the help of following steps: 1) 2) 3) 4) 5) 6) Import the data Select the MegaStat from Add-Ins option Select the Compare Independent Two Proportions from Hypothesis Tests Select data input Select Confidence level, Test difference and alternative hypothesis. Click Ok. Acti

Hypothesis test for two independent proportions p1 p2 0.2655 p (as decimal) 388/1223 544/2287 932/3510 p (as fraction) 0.317 0.238 544.306 387.691 931.997 X 2287 1223 3510 n 0.079 difference 0. hypothesized difference 0.0156 std. error 5.05 z 4.42E-07 p-value (two-tailed) From the Excel output, the z-test statistics is 5.05 The p-value for this test is, From the Excel output, the p-value for this test is 0.0000 Decision

The conclusion is that the p-value in this context is less than 0.05 which is 0.0000, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that there is a significant difference between inactive exercise rates and napping. The result is statistically significant