Find each percent of change. Round answers to the nearest tenth of a percent is necessary. 76 is decreased to 55
1 answer:
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Answer:
Step-by-step explanation:
Hello!
a)
One of the advantages of assigning the units randomly to the different treatments is to avoid selection bias. An example of selection bias would be selecting the best subjects at the barbell curl to the "encouraged" group and the worst ones to the "not encouraged" group.
Randomization also guarantees independence between the experimental units and treatment groups.
b) (see boxplots in attachment)
Encouraged group:
Box:
Q₁≅ 10
Q₂≅ 30
Q₃= 50
IQD= 50 - 10 = 40
The box seems symmetric, the distance between Q₁-Q₂ and Q₂-Q₃ is the same and the median is exactly in the middle.
Whiskers:
The left whisker is shorter than the right whisker, which is longer.
If you were to look only at the box, the distribution seems symmetrical but adding the whiskers there is a pronounced right asymmetry.
Min value ≅ 5
Max value ≅ 75
Range= 75 - 5 = 70
Not encouraged group:
Box:
Q₁≅ 10
Q₂≅ 20
Q₃= 35
IQD= 35 - 10 = 25
The box shows a little skewness to the right (the 2nd Quartile is closer to the first one than the 3rd) and is overall shorter compared to the first group.
Whiskers:
The left whisker is shorter than the right whisker, so in general, this distribution is also right-skewed.
Min value ≅ 0
Max value ≅ 60
Range= 60 - 0 = 60
Compared with the "encouraged" group, the IQD and the Range are shorter, which means that the "not encouraged" group shows less dispersion.
c) and d)
Daniel claims that people perform better in the barbell curl, on average, when being encouraged.
He divided the 29 subjects into two groups and measured their performance, then the study variables are:
X₁: Performance in the barber curl of a subject that was encouraged by his coach.
X₂: Performance in the barber curl of a subject that did not receive encouragement.
Logically, since Daniel wants to study the performance of both groups on average, you'd think that the parameters of interest will be the population means. But to study the average you need the population to have a normal distribution and looking at those boxplots, none of them is near the normal distribution and the size of both samples isn't big enough for an approximation.
The best option is to conduct a non-parametric analysis, for example, the Mann Whitney U-test and instead of comparing the population means, you'll compare the population medians: θ₁ vs. θ₂ ⇒ In the hypothesis of this test, you'll state that both samples come from the same population and if both have the same median, then each observation of the first sample x₁i will have an equal probability of being smaller or greater than each observation of the second sample x₂i (probability 0.5)
Depending of the statistics course, the hypotheses may change, I'm used at working directly with the medians, if they are equal, it will mean that the top 50% and bottom 50% of each population will be the same, which is the same as saying that P(x₁i > x₂i)= 0.5 vs. P(x₁i > x₂i)≠ 0.5
The conditions for the Mann-Whitney U-test are:
1) All observations on both groups should be independent. Check
2) The variables should be continuous or at least ordinal. Check
3) Both variables have the same distribution under the null hypothesis. Check (Looking at the boxplots, both are right-skewed)
4) Under the alternative hypothesis, the values of one of the populations exceed the other. In this case: H₁: θ₁ > θ₂
e)
H₀: θ₁ ≤ θ₂ vs. H₁: θ₁ > θ₂
Using the p-value approach, the decision rule is always the same:
If p-value ≤ α ⇒ Reject the null hypothesis.
If p-value > α ⇒ Do not reject the null hypothesis.
p-value: 0.107 > α: 0.05 ⇒ The decision is to not reject the null hypothesis.
At a 5% significance level, there is no significant evidence to reject the null hypothesis. Both samples seem to be from the same population. He can conclude that encouraging the subjects doesn't change significantly their performance in the barbell curl.
I hope you have a SUPER day!
