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,
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,
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter4: Equations Of Linear Functions
Section4.5: Correlation And Causation
Problem 11PPS
Related questions
Question
Answer the blanks in red. For the test procedure and p value use image 2
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff78bfed9-e56a-47b5-ad27-1a4a6c635001%2F2e956982-cc54-4074-ab39-299eaa246fbc%2Fomfzipa_processed.jpeg&w=3840&q=75)
Transcribed Image 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
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,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff78bfed9-e56a-47b5-ad27-1a4a6c635001%2F2e956982-cc54-4074-ab39-299eaa246fbc%2F7ih85wf_processed.jpeg&w=3840&q=75)
Transcribed Image 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,
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