3. BROILER DIET STUDY Some statistics With phytase: mean = 0.0637; median = 0.0635 Without phytase: mean = 0.0549; median = 0.0550 Let difference = with – without d = 0.0088 md = 0.009 Wilk-Shapiro...

Answer the following with the red hightlights. Use the second image to determine the test procedure and p-value3. BROILER DIET STUDY<br>Some statistics<br>With phytase:<br>mean = 0.0637; median = 0.0635<br>Without phytase: mean = 0.0549; median = 0.0550<br>Let difference = with – without<br>d = 0.0088<br>md = 0.009<br>Wilk-Shapiro Test for Normality<br>data = with<br>data = without<br>data:<br>difference<br>W = 0.98311<br>W = 0.93815<br>p-value = 0.5326<br>W = 0.89879<br>p-value = 0.2125<br>p-value = 0.9796<br>Student's t-test on Paired Samples: With vs Without<br>Ha: less than 0<br>Ha: not equal to 0<br>t = 6.527<br>Ha: greater than 0<br>t = 6.527<br>t = 6.527<br>p-value = 0.000108<br>p-value = 0.9999<br>p-value = 0.00005401<br>Wilcoxon Matched-Pairs Signed Ranks Test: With vs Without<br>Ha: not equal to 0<br>V = 55<br>Ha: less than 0<br>V = 55<br>p-value = 0.9979<br>Ha: greater than 0<br>V = 55<br>p-value = 0.005857<br>p-value = 0.002929<br>F-test for Equality of Variances<br>F = 0.89873<br>p-value = 0.8762<br>Student's t-test on Two Independent Pop'n Means: With vs Without<br>Ha: less than 0<br>Ha: not equal to 0<br>t = 6.4992<br>p-value= 0.000004121<br>Ha: greater than 0<br>t = 6.4992<br>p-value = 0.00000206<br>t = 6.4992<br>p-value = 1<br>Welch's t-test on Two Independent Pop'n Means: With vs Without<br>Ha: less than 0<br>Ha: not equal to 0<br>t = 6.4992<br>p-value= 0.000004179<br>Ha: greater than 0<br>t = 6.4992<br>p-value = 0.000002089<br>t = 6.4992<br>p-value = 1<br>Mann-Whitney Test on Two Independent Pop'n Means: With vs Without<br>Ha: not equal to 0<br>W = 99<br>Ha: less than 0<br>W = 94<br>Ha: greater than 0<br>W = 94<br>p-value = 0.0002396<br>p-value = 0.9999<br>p-value<br>= 0.0001198<br>

Extracted text: 3. BROILER DIET STUDY Some statistics With phytase: mean = 0.0637; median = 0.0635 Without phytase: mean = 0.0549; median = 0.0550 Let difference = with – without d = 0.0088 md = 0.009 Wilk-Shapiro Test for Normality data = with data = without data: difference W = 0.98311 W = 0.93815 p-value = 0.5326 W = 0.89879 p-value = 0.2125 p-value = 0.9796 Student's t-test on Paired Samples: With vs Without Ha: less than 0 Ha: not equal to 0 t = 6.527 Ha: greater than 0 t = 6.527 t = 6.527 p-value = 0.000108 p-value = 0.9999 p-value = 0.00005401 Wilcoxon Matched-Pairs Signed Ranks Test: With vs Without Ha: not equal to 0 V = 55 Ha: less than 0 V = 55 p-value = 0.9979 Ha: greater than 0 V = 55 p-value = 0.005857 p-value = 0.002929 F-test for Equality of Variances F = 0.89873 p-value = 0.8762 Student's t-test on Two Independent Pop'n Means: With vs Without Ha: less than 0 Ha: not equal to 0 t = 6.4992 p-value= 0.000004121 Ha: greater than 0 t = 6.4992 p-value = 0.00000206 t = 6.4992 p-value = 1 Welch's t-test on Two Independent Pop'n Means: With vs Without Ha: less than 0 Ha: not equal to 0 t = 6.4992 p-value= 0.000004179 Ha: greater than 0 t = 6.4992 p-value = 0.000002089 t = 6.4992 p-value = 1 Mann-Whitney Test on Two Independent Pop'n Means: With vs Without Ha: not equal to 0 W = 99 Ha: less than 0 W = 94 Ha: greater than 0 W = 94 p-value = 0.0002396 p-value = 0.9999 p-value = 0.0001198
3. BROILER DIET STUDY<br>A study aims to determine if a diet supplemented with phytase has a greater effect from<br>the diet with no phytase on the weight gain (kg) of the broilers. A random sample of 10<br>broilers were fed with the diet supplemented with phytase and another random sample of<br>10 broilers were fed with the diet without phytase. After a week, the weight of all the broilers<br>were collected, then the weight gain (kg) was computed.<br>(Use diet with phytase - diet without phytase in your computations)<br>Provide answers to the following:<br>a. At 5% level of significance, test if the diet supplemented with phytase has a<br>greater effect from the diet with no phytase on the weight gain (kg) of the<br>broilers.<br>Ho: (in symbols)<br>with the diet supplemented with phytase is<br>gain of broilers fed with the diet without phytase.<br>(in words) The average weight gain of broilers fed<br>the average weight<br>Ha: (in symbols)<br>with the diet supplemented with phytase is<br>gain of broilers fed with the diet without phytase.<br>(in words) The average weight gain of broilers fed<br>the average weight<br>Test Procedure:<br>Justify your choice of test procedure:<br>p-value:<br>Conclusion: At alpha = 0.05,<br>

Extracted text: 3. BROILER DIET STUDY A study aims to determine if a diet supplemented with phytase has a greater effect from the diet with no phytase on the weight gain (kg) of the broilers. A random sample of 10 broilers were fed with the diet supplemented with phytase and another random sample of 10 broilers were fed with the diet without phytase. After a week, the weight of all the broilers were collected, then the weight gain (kg) was computed. (Use diet with phytase - diet without phytase in your computations) Provide answers to the following: a. At 5% level of significance, test if the diet supplemented with phytase has a greater effect from the diet with no phytase on the weight gain (kg) of the broilers. Ho: (in symbols) with the diet supplemented with phytase is gain of broilers fed with the diet without phytase. (in words) The average weight gain of broilers fed the average weight Ha: (in symbols) with the diet supplemented with phytase is gain of broilers fed with the diet without phytase. (in words) The average weight gain of broilers fed the average weight Test Procedure: Justify your choice of test procedure: p-value: Conclusion: At alpha = 0.05,
Jun 07, 2022
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