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muminat
4 years ago
7

A study of 10 different weight loss programs involved 500 subjects. Each of the 10 programs had 50 subjects in it. The subjects

were followed for 12 months. Weight change for each subject was recorded. The researcher wants to test the claim that all ten programs are equally effective in weight loss. (a) Which statistical approach should be used? (i) confidence interval (ii) t-test (iii) ANOVA (iv) Chi square (b) Explain the rationale for your selection in (a). Specifically, why would this be the appropriate statistical approach?
Mathematics
1 answer:
Yakvenalex [24]4 years ago
8 0

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 p=10 groups and on each group from j=1,\dots,p=10 we have n_j individuals on each group we can define the following formulas of variation:  

SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2

SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2

SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2

And we have this property

SST=SS_{between}+SS_{within}

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.

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What is the solution to 3(–3x + 9) = −18?
Rus_ich [418]

Answer:

x = 5 (If solved algebraically)

No Solution (If the parentheses are absolute value lines)

Step-by-step explanation:

Result when solving this way: 3|-3x + 9| = -18

1) Divide both sides by 3
|-3x + 9| = -18/3

2) Simplify 18/3 to 6

|-3x + 9| = -6

3)  Break down the problem into these 2 equations.

-3x + 9 = -6

- (-3x + 9) = -6

4) Solve the 1st equation: -3x + 9 = -6

- Subtract 1 from both sides.

-3x = -6 - 9

-Simplify -6 - 9 to -15.

-3x = -15

-Divide both sides by -3.

x = -15/-3

- Two negatives make a positive.

x = 15/3

- Simplify 15/3 to 5.

x = 5

6)  Solve the 2nd equation:  - (-3x+ 9) = -6

1 - Remove parentheses

3x - 9 = -6

2 - Add 9 to both sides

3x = -6 + 9

3 - Simplify -6 + 9 to 3.

3x = 3

- Divide both sides by 3.

x = 1

6) Collect all solutions.

x = 1,5

7) Check solution.

When x = 1, the original equation 3|-3x + 9| = -18 does not hold true.

We will drop x = 1 from the solution set.

8) Check solution.

When x = 5, the original equation 3|-3x + 9| = -18  does not hold true.

We will drop x = 5 from the solution set.

9) Therefore,

No solution exists.

Result when solving algebraically:

Simplifying

3(-3x + 9) = -18

Reorder the terms:

3(9 + -3x) = -18

(9 * 3 + -3x * 3) = -18

(27 + -9x) = -18

Solving

27 + -9x = -18

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-27' to each side of the equation.

27 + -27 + -9x = -18 + -27

Combine like terms: 27 + -27 = 0

0 + -9x = -18 + -27

-9x = -18 + -27

Combine like terms: -18 + -27 = -45

-9x = -45

Divide each side by '-9'.

x = 5

Simplifying

x = 5

5 0
2 years ago
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