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from skin You are performing statistical analysis for a study where a biopsy is removed implants of young and older adults treated for burns. This study will aim to determine differences in rejection of tissues across these age groups. For each of the different iterations below, identify the best test. Describe the limitations for using the other tests described. (4 points each) 4. Experiment 1: A. Outcome sample space: (rejected, not rejected) B. Groups: Young, older C. Contingency table: Rejected Not rejected Young OlderExperiment 2: A. Outcome sample space: (rejected, partially rejected, not rejected) B. Groups: Young, older C. Contingency table: Rejected Partially rejected Not rejected Young Older 104122 156 14 123 Experiment 3: A. Outcome sample space (rejected, not rejected) B. Groups: Young, older C. Other rationale: Strong evidence exists that suggests that older adults tend to have reduced immune responses, which reduces the rejection of foreign tissues. D. Contingency table: Rejected 297 248 Not rejected Young Older 609 723 Perform the appropriate statistical analysis for each of the above data sets (using your answer in number 4). Specify the nullDand alternative hypothèses, perform the test, report the p- value, and report the conclusions of your statistical analysis. Use an alpha value of 0.05. (5 points each) 5. n er eaive hypeonse datases

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1 Chi-Square Test - Experiment 1 Observed Frequencies Tissues Calculations fo-fe 4 Age groups Rejected Not rejected Total Young 1.826091.826087 Older Total Expected Frequencies 15 1.8260871.82609 23 Tissues (fo-fe)A2/fe 0.871542 0.798913 0.464822 0.426087 10 Age grou Rejected Not rejected Total Young3.8260869574.173913043 Older 7.173913043 7.826086957 12 12 15 Total 23 Data 15 Level of Significance 16 Number of Rows 17 Number of Columns 18 Degrees of Freedom 19 20 Critical Value 21 Chi-Square Test Statistic |2.561363636 the proportions are same because the p-value of 0.1095 is 22 p-Value 23 Do not reject the null hypothesis significant difference in the two population proportions 0.05 2 2 In this experiment 1, the expected frequency is less than 5 1 for the young age groups for the expected rejected and not Results rejected frequencies less than 5. So this is not a good fit 3.841458821 model. Here we do not reject null hypothesis of the both 0.109504093 greater than the level of significance of the test. There is no 25 Expected frequency assumption 26 is violated.

Chi-Square Test - Experiment 1 Observed Frequencies Tissues Calculation 4 Age groups Rejected Not rejected Total fo-fe -SUM(B5:C5) SUM(B6:C6 -SUM(B7:C7 Young [2 B5-B11 C5-C11 B6-B12 C6-C12 Older 9 Total-SUM(B5:B6) -SUM(C5:C6) Expected Frequencies 10-A4 11 EA5 12-A6 13 B3 B4 SD5*B$7/SDS7 (fo-fe)A2/fe F5A2/B11-G5A2/C11 F6A2/B12G6 2/C12 Total SD5°C$7/SD$7 SUM(B11:C11 ŞD6*C$7/SD$7 SUM(B12:C12 -SUM(C11:C12) SUM(B13:C13 Total-SUM(B11:B12 Data 15 Level of Significance 0.05 16 Number of Rows 17 Number of Columns 2 18 Degrees of Freedom 19 20 Critical Value 21 Chi-Square Test Statistic SUM(ŞF$11:$G$12 22 p-Value 23 IF(B22<B15,Reject the null hypothesis,Do not re In this experiment 1, the expected frequency is less than 5 for the for the expected rejected and not rejected frequencies less than 5 good fit model. Here we do not reject null hypothesis of the both t same because the p-value of 0.1095 is greater than the level of sign test. There is no significant difference in the two population propo (SBS16-1)*(ŠBS17-1 Results -CHIINV(B15,B18 -CHIDIST(B21,B18 25 Expected frequency assu 26=IF(OR(B11く5,C11<5,B12c5,C12<5), İs violated., is met.)1 Chi-Square Test - Experiment 2 Observed Frequencies Tissues Calculations Rejected Partially rejected Not reiected Total fo-fe 5.497227 -0.05915 -5.43808 5.49723 0.059155.438078 4 Age groups Young Old Total 104 123 227 248 293 541 14 156 278 Expected Frequencies Tissues (fo-fe)A2/fe 1.831178 3.36E-05 0.232055 1.549939 2.85E-05 0.196415 Age groups Reiected Partially reiected Not reiected Total 12 13 Young 16.50277264 Old| 19.49722736 36 104.0591497 127.4380776 122.9408503 150.5619224 278 248 293 541 Total 227 Data 15 Level of Significance 16 Number of Rows 17 Number of Columns 18 Degrees of Freedom 19 20 Critical Value 21 Chi-Square Test Statistic 3.809649322 reject the null hypothesis. And conclude that all the 22 p-Value 23 Do not reject the null hypothesis model to the research question 0.05 2 3 Since p-value-0.14885 is greater than the level of test 2 significance of alpha 0.05, and the chi-square test Results statistic is less than the critical value of the chi-squared 5.991464547 distribution with 2 degrees of freedom, so we do not 0.148848739 three proportions are same. This is also not a good fitting 25 Expected frequency assumption 26 is met.1 Chi-Square Test - Expe Observed Frequencies Tissues Calculations Rejected Partially rejected Not rejected 122 156 Total fo-fe -C5-C11 -C6-C12 4 Age groups -B5-B11 -D5-D11 Young 22 Old 14 104 123 -SUM(C5:C6) :SUM(B5:D5 -SUM (B6:D6 -SUM(B7:D7) -B6-B12 =D6-D12 Total-SUM(B5:B6) Expected Frequencies B3 B4 SE5*BS7/SES7 SE6*BS7/SES7 Total (fo-fe)A2/fe 10-A4 11 EA5 12-A6 13 SE5*CS7/SES7 SE6*CS7/SES7 -SUM(C11:C12) D4 ŞE5 D$7/ŞE$7-SUM(B11:D11) SE6*DS7/SES7-SUM(B12:D12) -G6A2/B12 H6 2/C12 16 2/D12 Total-SUM(B11:B12 -SUM(D11:D12) SUM(B13:D13) Data 15 Level of Significance 0.05 16 Number of Rows 17 Number of Columns 3 18 Degrees of Freedom |=(SBS16-1)*(SBS17-1) | and the chi-square test statistic is less than the critical value of the chi-squared 19 20 Critical Value 21 Chi-Square Test Statis SUM(SG$11:$I$12) model to the research question. 22 p-Value 23 IF(B22<B15,Reject the null hypothesis, Since p-value-0.14885 is greater than the level of test significance of alpha 0.05 Results distribution with 2 degrees of freedom, so we do not reject the null hypothesis. And conclude that all the three proportions are same. This is also not a good fitting -CHIINV(B15,B18 -CHIDIST(B21,B18 25 Expected frequency as 26-IF(OR(B11<1,C11<1,C1 Chi-Square Test - Experiment 3 Observed Frequencies Tissues Calculations fo-fe 4 Age groups Rejected Not rejected Total Younger Old Total 297 248 545 906 971 1877 33.936633.9366 723 1332 33.9366 33.9366 Expected Frequencies Tissues (fo-fe)A2/fe 4.378005 1.791301 4.084936 1.671389 10 Age grou Rejected Not rejected Total Younger! 263.063399| 642.936601 Old 281.936601689.063399 1332 906 971 1877 Total Data 0.05 Null hypothesis: 15 Level of Significance 16 Number of Rows 17 Number of Columns 18 Degrees of Freedom 19 20 21 Critical Value 22 Chi-Square Test Statistic 11.92563044 population proportions. The experiment 3 is the good fitting model for the test 23 p-Value Ho 2 π.-To No difference between two population proportions Alternative hypothesis: 2 H1 TT2 Two population proportions are not same Since the chi-square test statistic,11.9256 is greater than critical value of 3.8415 and p-value is less the level of significance of the test, alpha-0.05, we reject the Results 3.841458821 null hypothesis, and conclude that there is a significant difference the two 0.000553669 in question of hypothesis testing Reject the null hypothesis 25 Expected frequency assumption 26 is met.1 Chi-Sqe Test- Experiment Observed Frequencies Tissues Calculations Rejected Not rejected Total fo-fe Younger 297 Old 248 SUM(B5:C5 -SUM(B6:C6) -SUM(B7:C7) B5-B11 -B6-B12 C5-C11 723 -SUM(C5:C6) Total SUM(BS:B6) Expected Frequencies -B3 fo-fe) 2/fe 10 EA4 11 EAS 12 EA6 13 SD5 B$7/SDS7 SD6BS7/SDST Total SUM(B11:C11) -SUM(B12:C12) -SUM(B13:C13) -F5A2/B11 G5 2/C11 F6 2/B12 2/C12 Total SUM(B11:B12 SUM(C11:C12 Data Null hypothesis 15 Level of Significance 16 Number of Rows 17 Number of Columns 18 Degrees of Freedom 0.05 ,- 2 No difference between two population proportions Alternative hypothesis: H, π Two population proportions are not same -(SB$16-1) (SB$17-1 Since the chi-square test statistic, 11.9256 is greater than critical value of 3.8415, and p-value is less the level of significance of the test alpha-0.05, we reject the null hypothesis, and conclude that there is a significant difference the two population proportions. The experiment 3 is the good fitting model for the test in question of hypothesis testing. Results 20 21 Critical Value 22 Chi-Square Test Statistic SUM(SF$11:$G$12) 23 p-Value 24IF(B23 B15, Reject the null 25 Expected frequency assumptic -CHIINV(B15,B18 -CHIDIST(B22,B18) hesis, Do not rejectt

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