Help ASAP. Identify at least two ordered pairs to the solutions of the system

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tiny-mole [99]

I think the answer 0.17 I'm sorry I may be wronh

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3 years ago

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A study randomly assigned adult subjects to one of three exercise treatments: (1) a single long exercise period five days per we

Katen [24]

Answer:

Step-by-step explanation:

Hello!

For this study adult subjects were randomly assigned to three different exercise treatments establishing three different groups:

Group 1: Single long exercise period five days per week

Group 2: Several 10' exercise periods five days per week

Group 3: Several 10' exercise periods five days per week using a home treadmill

All subjects were weighed before and after 6 months of training and their weight loss was determined.

The objective of this study is to determine if there is any difference between the treatments, meaning, if the weight loss is the same regarding the type of exercise or if the type of exercise has some influence over it.

To test this you have to conduct an ANOVA, with the hypothesis:

H₀: μ₁= μ₂= μ₃

H₁: At least one μi differs from the others. ∀ i=1, 2, 3

α: 0.05

The statistic is:

$F= \frac{MS_{Tr}}{MS_{Er}} ~~F_{k-1; n-k}$

Normally it is better to have the raw data to conduct the analysis using statistical software, but it is not impossible to calculate the statistic using the descriptive statistics for each group.

As you know the statistic is calculated as the ratio between the mean square of the treatments (between groups) and the mean squares of errors.

Each means square is calculated as the sum of squares by the degrees of freedom of the category: SS/Df

So first you need to determine the SS and Df of treatments and errors, then the MS and finally the value of F.

Treatments:

> The degrees of freedom between treatments are k-1 (k represents the number of treatments):

DfTr: 3-1= 2

>The sum of squares is:

SSTr: ∑ni(Ÿi - Ÿ..)²

Ÿi= sample mean of sample i ∀ i= 1,2,3

Ÿ..= grand mean is the mean that results of all the groups together.

So the Sum of squares of treatments (SSTr) is the sum of the square of the difference between the sample mean of each group and the grand mean.

To calculate the grand mean you can sum the means of each group and dive it by the number of groups:

Ÿ..= (Ÿ₁ + Ÿ₂ + Ÿ₃)/ k = (10.2+9.3+10.2)/3= 29.7/3= 9.9

SSTr= (Ÿ₁ - Ÿ..)² + (Ÿ₂ - Ÿ..)² + (Ÿ₃ - Ÿ..)²= (10.2 - 9.9)² + (9.3 - 9.9)² + (10.2 - 9.9)²= 0.54

> The Means Squares is:

MSTr= SSTr/DFTr= 0.54/2= 0.27

Errors

>Degrees of freedom of error DfEr= N-k Where N is the total of observations of the experiment (N= n₁+n₂+n₃) and k is the number of treatments.

DfEr= N-k= 30-3= 27

>The mean square error (MSEr) is the estimation of the variance error (σ $_{e}^2$ → S $_{e}^2$ ), you have to use the following formula:

$S_{e}^2= \frac{(n_1-1)S^2_1+(n_2-1)S_2^2+(n_3-1)S^2_3}{N-k}= \frac{(9*17.64)+(9*20.25)+(9*27.04)}{27}= \frac{584.37}{27}= 21.643$

Finally you calculate the statistic:

$F_{H_0}= \frac{MS_{Tr}}{MS_{Er}}= \frac{0.27}{21.643}= 0.012$

This test is One-tailed right, meaning that you'll reject to high values of the statistic and there is one critical value:

$F_{k-1;N-k;1-\alpha }= F_{2;27;0.95}= 3.35$

Using the critical value approach, the decision rule is:

If $F_{H_0}$ ≥ 3.35, reject the null hypothesis.

If $F_{H_0}$ < 3.35, do not reject the null hypothesis.

The calculated value is less than the critical value, the decision is to not reject the null hypothesis.

So at a 5% significance level, there is not enough evidence to conclude that the treatments influence significantly in weight loss. The population mean weight loss is the same for the treatments "Single long exercise period five days per week", "Several 10' exercise periods five days per week" and "Several 10' exercise periods five days per week using a home treadmill"

I hope this helps!