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irakobra [83]
2 years ago
11

A study is designed to examine the effect of doing synchronized movements (such as marching in step or doing synchronized dance

steps) and the effect of exertion on many different variables, including how close participants feel to others in their group. In the study, high school students in Brazil were randomly assigned to an exercise with either high synchronization (HS) or low synchronization (LS) and also either to high exertion (HE) or low exertion (LE). Thus, there are four groups: HS+HE, HS+LE, LS+HE, and LS+LE. Closeness is measured on a 7-point Likert scale (1=least close to 7=most close), and the response variable is the change in how close participants feel to those in their group using the rating after the exercise minus the rating before the exercise.
The output for an ANOVA test is shown below, along with some summary statistics.

Analysis of Variance
Source DF SS MS F-Value P-Value
Group 3 27.04 9.012 2.77 0.042
Error 256 831.52 3.248
Total 259 858.55
Group N Mean StDev
HS+HE 72 0.319 1.852
HS+LE 64 0.328 1.861
LS+HE 66 0.379 1.838
LS+LE 58 −0.431 1.623

(a) In both groups with high synchronization (HS), does mean closeness rating go up or down after the synchronized exercise?
(b) In the groups with low synchronization (LS), does mean closeness rating go up or down if the group engages in high exertion (HE) exercise? How about if the group engages in low exertion (LE) exercise?
(c) How many students were included in the analysis?
(d) At a 5% level, what is the conclusion of the test? If there is a difference in means, indicate where the difference is likely to be.
(e) At a 1% level, what is the conclusion of the test?

Mathematics
1 answer:
Vlada [557]2 years ago
3 0

Answer:

The complete question in the graphic

A.

After synchronized exercise, it was observed that the average proximity rate has increased, which shows that the mean of the variable is positive and there is high synchronization in the groups.

.

B.

If groups with low synchronization and high effort are compared, the result of the average will be positive, so the average proximity rate will also increase after the synchronization exercise is performed.

On the contrary, if we observe the groups that have low synchronization and low effort, the mean of the variable is negative, thus the average proximity rate after performing the exercise will decrease.

(C)

considering that the total degrees of freedom is 259, it will show 260 students in the analysis.

D.

With a significance level of 5% or 0.05, the P value is less than the significance level, which leads to differentiating the mean of the scores before and after performing the combinations.

E.

With a significance level of 1% or 0.01, the p-value is higher than the significance level. With this data we can conclude that there is no difference between the results before and after the synchronized exercise

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Therefore, probability that the selected fish is a male green swordtail is \dfrac{9}{35}.

D.

We have to find the probability that the selected fish is either a male or a green swordtail.

\text{P(Male or green swordtail)}=\dfrac{\text{Total male or green swordtail fish}}{\text{Total fish}}

\text{P(Male or green swordtail)}=\dfrac{44+36+24}{140}

\text{P(Male or green swordtail)}=\dfrac{96}{140}

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Therefore, the probability the selected fish is either a male or a green swordtail is \dfrac{24}{35}.

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Answer:

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