homework. answer all questions.
Bios 1 Nurs 717AP HW #4 Question 1) Let say a researcher who wants to study on physical activity of nursing students. A researcher wants to examine the physical activity by student year in school (freshman, sophomore, junior, and senor). He will use established physical activity measurement. a) What statistical test researcher should use? 1. Two sample T-test 2. Analysis of Variance (ANOVA) 3. Two sample proportion 4. None of above b) Write the null and alternative hypothesis for part a. 1. Ho: 1 = 2 =3 = 4 and H1: At least one test is not equal 2. Ho: 1 = 2 and H1: At least one test is not equal 3. Ho: 3 = 4 and H1: At least one test is not equal 4. None of above Question 2) The following is a study to examine three methods (1-3) to reduce stress from patients with chronic disease. There were 5 subjects for each group. Use the output or calculate by hand. 1 2 3 Total 14 14 15 20 25 24 24 18 19 25 30 21 30 32 35 Total 88 110 148 346 Mean 17.6 22 29.60 23.07 Variance 23.3 10.5 27.3 43.78 n 5 5 5 15 Source DF Sum of Squares Mean Square F Value Pr > F Model 2 368.5333333 184.2666667 9.05 0.0040 Error 12 244.4000000 20.3666667 Corrected Total 14 612.9333333 a) What is the SSBetween? 1. 244.4 2. 100.55 3. 368.5 4. 442.36 b) What is the SSWithin or Error? 1. 244.4 2. 100.55 3. 197.96 4. 442.36 c) What is the SSTotal?. 1. 244.4 2. 100.55 3. 197.96 4. 612.9 d) What is the F value? 1. 9.05 2. 10.55 3. 19.96 4. 2.36 e) Can we conclude from these data the three methods differ in reducing stress using alpha .05? 1. Yes, Since F statistics is greater than F value from F table we can conclude from these data the three methods differ in reducing stress using alpha .05 2. No, Since F statistics is less than F value from F table we cannot conclude from these data the three methods differ in reducing stress using alpha .05 Output: Source DF Sum of Squares Mean Square F Value Pr > F Model 2 368.5333333 184.2666667 9.05 0.0040 Error 12 244.4000000 20.3666667 Corrected Total 14 612.9333333 Question 3) The following are SAS output for multiple comparison tests Tukey and Bonferroni Methods. Tukey Mehtod Alpha 0.05 Error Degrees of Freedom 12 Error Mean Square 20.36667 Critical Value of Studentized Range 3.77289 Minimum Significant Difference 7.6146 Comparisons significant at the 0.05 level are indicated by ***. x Comparison Difference Between Means Simultaneous 95% Confidence Limits group 3 - group 2 7.600 -0.015 15.215 group 3 - group 1 12.000 4.385 19.615 *** group 2 - group 3 -7.600 -15.215 0.015 group 2 - group 1 4.400 -3.215 12.015 group 1 - group 3 -12.000 -19.615 -4.385 *** group 1 - group 2 -4.400 -12.015 3.215 Bonferroni Method: Alpha 0.05 Error Degrees of Freedom 12 Error Mean Square 20.36667 Critical Value of t 2.77947 Minimum Significant Difference 7.9333 Means with the same letter are not significantly different. Bon Grouping Mean N x A 29.600 5 group 3 A B A 22.000 5 group 2 B B 17.600 5 group 1 What statement is correct for both tests? 1. Both methods show group 3 is only different from group1 2. Both methods did not show the same result 3. Tukey’s method indicated more significant than Bonferroni method. 4. Bonferroni method indicated more significant than Tukey method Question4) The effect of three types of physical activity and working status on reducing anxiety of 25 randomly selected of participant that comes for his or her annual check- up. The numbers of subjects in each clinic were classified by type of physical activity and working status as follow: Type of physical activity Working Status No Yes None 18,19,20,22 30,32,33,33,34 Light 35,36,37 20,21,22,23 Moderate to High 20,21,22,23,24 29,30,33,34 Use Sas Output to answer the following: a) What is the F value? 1. 14.864 2. 20.55 3. 15.96 4. 25.38 b) What effects are significant (the main effects and interaction)? Let alpha=0.05. 1. Only factor a 2. Only factor b 3. Only interaction effect a*b 4. Both factor a and b c) What is the result for adjusted Tukey for factor a? 1. Group light is different from moderate and none. Also, group moderate is different from none. 2. Group light is only different from none. 3. Group light is only different from moderate. 4. Group moderate is only different from none. d) Class Level Information Class Levels Values a 3 light moderate to high none b 2 No yes Number of Observations Read 25 Number of Observations Used 25 Source DF Sum of Squares Mean Square F Value Pr > F Model 5 559.2900000 111.8580000 25.38<.0001 error="" 19="" 83.7500000="" 4.4078947="" corrected="" total="" 24="" 643.0400000="" r-square="" coeff="" var="" root="" mse="" y mean="" 0.869759="" 11.84819="" 2.099499="" 17.72000="" source="" df="" type="" i="" ss="" mean="" square="" f="" value="" pr =""> F a 2 367.6114286 183.8057143 41.70<.0001 b="" 1="" 186.1270250="" 186.1270250="" 42.23=""><.0001 a*b="" 2="" 5.5515464="" 2.7757732="" 0.63="" 0.5435="" source="" df="" type="" iii="" ss="" mean="" square="" f="" value="" pr =""> F a 2 309.2010309 154.6005155 35.07<.0001 b="" 1="" 175.8353933="" 175.8353933="" 39.89=""><.0001 a*b="" 2="" 5.5515464="" 2.7757732="" 0.63="" 0.5435="" tukey="" multiple="" comparison:="" physical="" activity="" a="" y="" lsmean="" lsmean="" number="" light="" 18.0000000="" 1="" moderate="" to="" high="" 13.3500000="" 2="" none="" 21.6750000="" 3="" least squares means for effect a="" pr =""> |t| for H0: LSMean(i)=LSMean(j) Dependent Variable: y i/j 1 2 3 1 0.0009 0.0073 2 0.0009<.0001 3="" 0.0073=""><.0001 working="" status="" b="" y="" lsmean="" h0:lsmean1="LSMean2" pr =""> |t| No 14.9833333<.0001 yes="" 20.3666667="" question="" 5)="" the="" effect="" of="" three="" different="" treatments="" on="" depression="" used="" on="" 30="" randomly="" selected="" patients="" in="" one="" clinic.="" the="" numbers="" of="" treated="" patients="" in="" each="" clinic="" were="" classified="" by="" type="" of="" treatment="" and="" gender="" as="" follow:="" type="" of="" treatment="" gender="" male="" female="" treatment="" 1="" 18,19,20,22,23="" 30,32,33,33,34,35="" treatment="" 2="" 35,36,37,37="" 20,21,22,23="" treatment="" 3="" 20,21,22,23,24,25="" 29,30,33,34,35="" use="" sas="" output="" to="" answer="" the="" following:="" a)="" what="" is="" the="" f="" value?="" 1.="" 34.864="" 2.="" 64.70="" 3.="" 45.96="" 4.="" 55.38="" b)="" what="" effects="" are="" significant="" (the="" main="" effects="" and="" interaction)?="" let="" alpha="0.05." 1.="" only="" factor="" a="" 2.="" only="" factor="" b="" 3.="" only="" interaction="" effect="" a*b="" 4.="" factor="" a,="" b,="" and="" interaction="" effect="" c)="" how="" many="" significant="" differences="" you="" find="" for="" adjusted="" tukey="" for="" interaction="" effect?="" 1.="" 5="" 2.="" 8="" 3.="" 10="" 4.="" 3="" class="" level="" information="" class="" levels="" values="" a="" 3="" treatment="" 1="" treatment="" 2="" treatment="" 3="" b="" 2="" female="" male="" number="" of="" observations="" read="" 30="" number="" of="" observations="" used="" 30="" source="" df="" sum="" of="" squares="" mean="" square="" f="" value="" pr =""> F Model 5 1133.383333 226.676667 64.70<.0001 error="" 24="" 84.083333="" 3.503472="" corrected="" total="" 29="" 1217.466667="" r-square="" coeff="" var="" root="" mse="" y mean="" 0.930936="" 6.798147="" 1.871756="" 27.53333="" source="" df="" type="" i="" ss="" mean="" square="" f="" value="" pr =""> F a 2 20.0462121 10.0231061 2.86 0.0768 b 1 127.7829823 127.7829823 36.47<.0001 a*b="" 2="" 985.5541390="" 492.7770695="" 140.65=""><.0001 source="" df="" type="" iii="" ss="" mean="" square="" f="" value="" pr =""> F a 2 23.8800813 11.9400407 3.41 0.0498 b 1 44.2002252 44.2002252 12.62 0.0016 a*b 2 985.5541390 492.7770695 140.65<.0001 tukey="" method:="" treatment="" a="" y="" lsmean="" lsmean="" number="" treatment="" 1="" 26.6166667="" 1="" treatment="" 2="" 28.8750000="" 2="" treatment="" 3="" 27.3500000="" 3="" least squares means for effect a="" pr =""> |t| for H0: LSMean(i)=LSMean(j) Dependent Variable: y i/j 1 2 3 1 0.0409 0.6363 2 0.0409 0.2077 3 0.6363 0.2077 Gender b y LSMEAN H0:LSMean1=LSMean2 Pr > |t| female 28.8444444 0.0016 male 26.3833333 Interaction comparison a b y LSMEAN LSMEAN Number treatment 1 female 32.8333333 1 treatment 1 male 20.4000000 2 treatment 2 female 21.5000000 3 treatment 2 male 36.2500000 4 treatment 3 female 32.2000000 5 treatment 3 male 22.5000000 6 Least Squares Means for effect a*b Pr > |t| for H0: LSMean(i)=LSMean(j) Dependent Variable: y i/j 1 2 3 4 5 6 1<.0001><.0001 0.0869="" 0.9928=""><.0001 2=""><.0001 0.9486=""><.0001><.0001 0.4532="" 3=""><.0001 0.9486=""><.0001><.0001 0.9594="" 4="" 0.0869=""><.0001><.0001 0.0374=""><.0001 5="" 0.9928=""><.0001><.0001 0.0374=""><.0001 6=""><.0001 0.4532="" 0.9594=""><.0001><.0001 question 6) a sample of 1000 women who had children were asked about breast feeding and if they had breast cancer. the following table is the cross tables for breast cancer and breast feeding. calculate by hand or use the output to answer the questions. breast cancer breast feeding no yes total no 300 100 400 yes 550 50 600 total 850 150 1000 a) what are the expected values for cell1 and cell 4? 1. 340 and 90 2. 60 and 510 3. 60 and 90 4. 340 and 510 b) what is the chi square value? 1. 34.10 2. 6.90 3. 20.60 4. 52.29 c) what type of chi square test is this problem? 1. an independent test 2. homogenous test d) do these data indicate that there is association between breast feeding and breast cancer? let alpha=.01. 1. yes, since chi square test is greater than chi square value from table the data indicate that there is association between breast feeding and breast cancer. 2. no, since chi square test is less than chi square value from table question="" 6)="" a="" sample="" of="" 1000="" women="" who="" had="" children="" were="" asked="" about="" breast="" feeding="" and="" if="" they="" had="" breast="" cancer.="" the="" following="" table="" is="" the="" cross="" tables="" for="" breast="" cancer="" and="" breast="" feeding.="" calculate="" by="" hand="" or="" use="" the="" output="" to="" answer="" the="" questions.="" breast="" cancer="" breast="" feeding="" no="" yes="" total="" no="" 300="" 100="" 400="" yes="" 550="" 50="" 600="" total="" 850="" 150="" 1000="" a)="" what="" are="" the="" expected="" values="" for="" cell1="" and="" cell="" 4?="" 1.="" 340="" and="" 90="" 2.="" 60="" and="" 510="" 3.="" 60="" and="" 90="" 4.="" 340="" and="" 510="" b)="" what="" is="" the="" chi="" square="" value?="" 1.="" 34.10="" 2.="" 6.90="" 3.="" 20.60="" 4.="" 52.29="" c)="" what="" type="" of="" chi="" square="" test="" is="" this="" problem?="" 1.="" an="" independent="" test="" 2.="" homogenous="" test="" d)="" do="" these="" data="" indicate="" that="" there="" is="" association="" between="" breast="" feeding="" and="" breast="" cancer?="" let="" alpha=".01." 1.="" yes,="" since="" chi="" square="" test="" is="" greater="" than="" chi="" square="" value="" from="" table="" the="" data="" indicate="" that="" there="" is="" association="" between="" breast="" feeding="" and="" breast="" cancer.="" 2.="" no,="" since="" chi="" square="" test="" is="" less="" than="" chi="" square="" value="" from="">