Answer:
a) (iii) ANOVA
b) The ANOVA test is more powerful than the t test when we want to compare group of means.
Step-by-step explanation:
Previous concepts
Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".
The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"
If we assume that we have 
groups and on each group from 
we have 
individuals on each group we can define the following formulas of variation:
And we have this property
Solution to the problem
Part a
(i) confidence interval
False since the confidence interval work just when we have just on parameter of interest, but for this case we have more than 1.
(ii) t-test
Can be a possibility but is not the best method since every time that we conduct a t-test we have a chance that we commit a Type I error.
(iii) ANOVA
This one is the best method when we want to compare more than 1 group of means.
(iv) Chi square
False for this case we don't want to analyze independence or goodness of fit, so this one is not the correct test.
Part b
The ANOVA test is more powerful than the t test when we want to compare group of means.
Not too sure but try 18 sorry if it does not help
Answer:
Multiple answers
Step-by-step explanation:
Considering that 8.1% of Americans have the disease:
If our theoretical group is of 10,000. We know that the 8.1% of our group have diabetes, so we multiplify 10,000 × .081(the percentaje) = 810. This is the total of adults in our group that have diabetes.
Now:
- We know that the test correctly diagnoses 95% of adults with diabetes. In our case the total of adults with diabetes is 810, so we multiplify 810 × .95(percentaje) = 769.50.
- We know too that the test incorrectly diagnoses 3.5% of the adults. In our case the total of adults with diabetes is 810, so we multiplify 810 × .035(percentaje) = 28.35.
- The total of adults the test diagnoses positive with diabetes should be the correctly and incorrectly we calculate previously. 769.50 + 28.35 = 797.85
- The total of test negative diagnoses should be the total of our group less the positive diagnoses. 10,000 - 797.85 = 9,202.15
- Total do not have diabetes:
- The total of adults do not have diabetes is the total of our group less the total of adults in our group that have diabetes. 10,000 - 810 = 9190
We expect that only the 95% of test positive for diabetes have the disease.
- 810 × .95(percentaje) = 769.50.
We expect that only the 5% (100% - 95% of test positive) of the 8.1% of americans afflicted with diabetes of negative test actually have the disease. 8.1 × .05(percentaje) = .40%, 9,202.15 ×.004(percentaje) = 36.80.
The 3.5% of Americans who test positive will not have the disease because this is the percentaje that the test incorrectly diagnoses.
Answer:
41
Step-by-step explanation:
It increases then decreases then becomes constant
