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Damm [24]
2 years ago
10

Researchers are investigating whether people who exercise with a training partner have a greater increase, on average, in target

ed exercise intensity compared with people who exercise alone. Two methods of collecting data have been proposed.
ᄋ Method I: Recruit volunteers who are willing to participate. Randomly assign each participant to exercise with a training partner or to exercise alone.

ᄋ Method II: Select a random sample of people from all the people who exercise at a community fitness center. Ask each person in the sample whether they use a training partner, and use the response to create the two groups.

(a) For each method, the researchers will record the change in targeted exercise intensity for each person in the investigation. They will compare the mean change in intensity between those who exercise with a training partner and those who do not.


(i) Describe the population of generalization if method I is used.

(ii) Describe the population of generalization if method II is used.

(b) Suppose the investigation produces a result that is statistically significant using both methods. What can be concluded if method I is used that cannot be concluded if method II is used?
Mathematics
1 answer:
Sphinxa [80]2 years ago
7 0

Answer:

we have to compare the effects of method I and method II.

(a)

(i) The population of generalization if the method I is used will be all people who are willing to determine the effect of exercising with a training partner as compared to exercising alone.

(ii) The population of generalization, if method II is used, will be all people who exercise at a community fitness center.

(b)

If both the methods produce significant results then the difference between the conclusion of the method I and the conclusion of method II is that the result of the method I can be generalized to a broader population since it does not restrict people to participate in the study.

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Serhud [2]

Answer:

Option A

Step-by-step explanation:

A type I error is committed when a researcher rejects the null hypothesis when it is actually true.

The null hypothesis is: U <= 50%

The alternative is: U > 50%

Thus, the principal could have committed an error by rejecting null hypothesis and concluding that more than 50% of students want earlier lunch, when in actuality 50% or less want earlier lunch.

5 0
2 years ago
You and your team are planning a space mission. You want to visit each planet listed in the chart in the most efficient way poss
Daniel [21]
You would go to Neptune, Uranus, Saturn, Jupiter, Mars, Earth (to refuel), Mercury, Venus
6 0
2 years ago
Jacque needs to buy some pizzas for a party at her office. She's ordering from a restaurant that charges a \$7.50$7.50dollar sig
BigorU [14]

Answer:

Step-by-step explanation:

Delivery fee = $7.50

Price per pizza = $14

Planned amount = $60

7.50 + 14p < 60

Subtract 7.50 from both sides

14p < 60 - 7.50

14p < 52.50

Divide both sides by 14

p < 52.50 / 14

p < 3.75

Each pizza is cut into 8 slices

Total slices she can afford = 3.75 × 8

= 30 slices

5 0
2 years ago
Compare the values of the 2s and 5s in 55,220
bagirrra123 [75]
The five digit number, 55,220 contains to 5's and two 2's.  The 5's are in the ten thousand and one thousand columns. This means that one five represents 50,000 and the second represents 5, 000.  The two's are in the hundreds and tens columns meaning one 2 represents 200 and the other 2 represents 20. In a direct comparison of these numbers the 5's equal 55,000 and the 2's equal 220.
6 0
2 years ago
Three assembly lines are used to produce a certain component for an airliner. To examine the production rate, a random sample of
nikitadnepr [17]

Answer:

a) Reject H₀

b) [0.31; 3.35]

Step-by-step explanation:

Hello!

a) The objective of this example is to compare if the population means of the production rate of the assembly lines A, B and C. To do so the data of the production of each line were recorded and an ANOVA was run using it.

The study variable is:

Y: Production rate of an assembly line.

Assuming that the study variable has a normal distribution for each population, the observations are independent and the population variances are equal, you can apply a parametric ANOVA with the hypothesis:

H₀ μ₁= μ₂= μ₃

H₁: At least one of the population means is different from the others

Where:

Population 1: line A

Population 2: line B

Population 3: line C

α: 0.01

This test is always one-tailed to the right. The statistic is the Snedecor's F, constructed as the MSTr divided by the MSEr if the value of the statistic is big, this means that there is a greater variance due to the treatments than to the error, this means that the population means are different. If the value of F is small, it means that the differences between populations are not significant ( may differ due to error and not treatment).

The critical region is:

F_{k-1;n-k; 1-\alpha } = F_{2;15; 0.99} = 6.36

If F ≥ 3.36, the decision is to reject the null hypothesis.

Looking at the given data:

F= \frac{MSTr}{MSEr}= 11.32653

With this value the decision is to reject the null hypothesis.

Using the p-value method:

p-value: 0.001005

α: 0.01

The p-value is less than the significance level, the decision is to reject the null hypothesis.

At a level of 5%, there is significant evidence to say that at least one of the population means of the production ratio of the assembly lines A, B and C is different than the others.

b) In this item, you have to stop paying attention to the production ratio of the assembly line A to compare the population means of the production ratio of lines B and C.

(I'll use the same subscripts to be congruent with part a.)

The parameter to estimate is μ₂ - μ₃

The populations are the same as before, so you can still assume that the study variables have a normal distribution and their population variances are unknown but equal. The statistic to use under these conditions, since the sample sizes are 6 for both assembly lines, is a pooled-t for two independent variables with unknown but equal population variances.

t=  (X[bar]₂ - X[bar]₃) - ( μ₂ - μ₃) ~t_{n_2+n_3-2}

Sa√(1/n₂+1/n₃)

The formula for the interval is:

(X[bar]₂ - X[bar]₃) ± t_{n_2+n_3-2; 1 - \alpha /2}* Sa\sqrt{*\frac{1}{n_2} + \frac{1}{n_3} }

Sa^{2} = \frac{(n_2-1)*S_2^2+ (n_3-1)*S_3^2}{n_2+n_3-2}

Sa^{2} = \frac{(5*0.67)+ (5*0.7)}{6+6-2}

Sa^{2} = 0.685

Sa= 0.827 ≅ 0.83

t_{n_2+n_3-2;1-\alpha /2}= t_{10;0.995} = 3.169

X[bar]₂ = 43.33

X[bar]₃ = 41.5

(43.33-41.5) ± 3.169 * *0.83\sqrt{*\frac{1}{6} + \frac{1}{6} }

1.83 ± 3.169 * 0.479

[0.31; 3.35]

With a confidence level of 99% you'd expect that the difference of the population means of the production rate of the assemly lines B and C.

I hope it helps!

8 0
2 years ago
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