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Troyanec [42]
3 years ago
15

To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been randomly assig

ned to the 3 treatments. You are given the results below.
Treatment | Observation
A | 20 | 30 | 25 | 33
B | 22 | 26 | 20 | 28
C | 40 | 30 | 28 | 22

1.The null hypothesis for this ANOVA problem is?

2.The mean square between treatments (MSTR) equals:
A. 1.872
B. 5.86
C.34
D.36

3.The mean square within treatments (MSE) equals:
A.1.872
B. 5.86
C. 34
D.36

4. The test statistic to test the null hypothesis equals:
A. .944
B.1.059
C. 3.13
D. 19.231

5. The null hypothesis is to be tested at the 1% level of significance. The critical value from the table is
A.4.26
B.8.02
C. 16.69
D. 99.39
Mathematics
1 answer:
Keith_Richards [23]3 years ago
5 0

Answer:

1. Null hypothesis: \mu_{A}=\mu_{B}=\mu_{C}

Alternative hypothesis: Not all the means are equal \mu_{i}\neq \mu_{j}, i,j=A,B,C

2. D. 36

3. C. 34

4. B. 1.059

5. B. 8.02

Step-by-step explanation:

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"

Part 1

The hypothesis for this case are:

Null hypothesis: \mu_{A}=\mu_{B}=\mu_{C}

Alternative hypothesis: Not all the means are equal \mu_{i}\neq \mu_{j}, i,j=A,B,C

Part 2

In order to find the mean square between treatments (MSTR), we need to find first the sum of squares and the degrees of freedom.

If we assume that we have p groups and on each group from j=1,\dots,p 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}

We need to find the mean for each group first and the grand mean.

\bar X =\frac{\sum_{i=1}^n x_i}{n}

If we apply the before formula we can find the mean for each group

\bar X_A = 27, \bar X_B = 24, \bar X_C = 30. And the grand mean \bar X = 27

Now we can find the sum of squares between:

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

Each group have a sample size of 4 so then n_j =4

SS_{between}=SS_{model}=4(27-27)^2 +4(24-27)^2 +4(30-27)^2=72

The degrees of freedom for the variation Between is given by df_{between}=k-1=3-1=2, Where  k the number of groups k=3.

Now we can find the mean square between treatments (MSTR) we just need to use this formula:

MSTR=\frac{SS_{between}}{k-1}=\frac{72}{2}=36

D. 36

Part 3

For the mean square within treatments value first we need to find the sum of squares within and the degrees of freedom.

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

SS_{error}=(20-27)^2 +(30-27)^2 +(25-27)^2 +(33-27)^2 +(22-24)^2 +(26-24)^2 +(20-24)^2 +(28-24)^2 +(40-30)^2 +(30-30)^2 +(28-30)^2 +(22-30)^2 =306

And the degrees of freedom are given by:

df_{within}=N-k =3*4 -3 = 12-3=9. N represent the total number of individuals we have 3 groups each one with a size of 4 individuals. And k the number of groups k=3.

And now we can find the mean square within treatments:

MSE=\frac{SS_{within}}{N-k}=\frac{306}{9}=34

C. 34

Part 4

The test statistic F is given by this formula:

F=\frac{MSTR}{MSE}=\frac{36}{34}=1.059

B. 1.059

Part 5

The critical value is from a F distribution with degrees of freedom in the numerator of 2 and on the denominator of 9 such that we have 0.01 of the area in the distribution on the right.

And we can use excel to find this critical value with this function:

"=F.INV(1-0.01,2,9)"

And we will see that the critical value is F_{crit}=8.02

B. 8.02

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Let <em>P(A) </em>be the probability that goggle of type A is manufactured

<em>P(B) </em>be the probability that goggle of type B is manufactured

<em>P(E)</em> be the probability that a goggle is returned within 10 days of its purchase.

According to the question,

<em>P(A)</em> = 30%

<em>P(B)</em> = 70%

<em>P(E/A)</em> is the probability that a goggle is returned within 10 days of its purchase given that it was of type A.

P(E/B) is the probability that a goggle is returned within 10 days of its purchase given that it was of type B.

P(A \cap E) will be the probability that a goggle is of type A and is returned within 10 days of its purchase.

P(B \cap E) will be the probability that a goggle is of type B and is returned within 10 days of its purchase.

P(E \cap A) = P(A) \times P(E/A)

P(E \cap A) = \dfrac{30}{100} \times \dfrac{5}{100}\\\Rightarrow P(E \cap A) = 1.5 \%

P(E \cap B) = P(B) \times P(E/B)

P(E \cap B) = \dfrac{70}{100} \times \dfrac{2}{100}\\\Rightarrow P(E \cap B) = 1.4 \%

P(E) = 1.5 \% + 1.4 \% \\P(E) = 2.9\%

If a goggle is returned within 10 days of its purchase, probability that it was of type B:

P(B/E) = \dfrac{P(E \cap B)}{P(E)}

\Rightarrow \dfrac{1.4 \%}{2.9\%}\\\Rightarrow \dfrac{14}{29}

So, the required probability is \dfrac{14}{29}.

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