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1 year ago

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Henry is 3 times older than Julia. Rhianna is 1/2 as old as Julia. What formula shows the relationship between Henry and Rhianna

's ages?
A) H = 6R
B) 3H = 2R
C) H =
3
2
R
D) 3H =
1
2
R

1 answer:

Sedbober [7]1 year ago

8 0

The answer is A because if you make up an age for Julia for example Julia is 6 years old. In that case we multiply her age by 3 to figure out Henry's age which is 18. Rhianna is half as old as Julia is so that would make her 3 years old. In order for Henry which he is 18 years old to get to Rhianna's age which is 3 years old. you just divide 18 by 3 which is 6 years old. therefore the answer is A

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

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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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Answer:

The 95% of confidence intervals

(2.84 ,2.99)

Step-by-step explanation:

A random sample of 20 accounting students results in a mean of 2.92 and a standard deviation of 0.16

given small sample size n =20

sample mean x⁻ =2.92

sample standard deviation 'S' =0.16

level of significance ∝ = 0.95

The 95% of confidence intervals

$x^{-} ± t_{\alpha } \frac{S}{\sqrt{n} }$

the degrees of freedom γ=n-1 =20-1=19

t-table 2.093

$(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} },x^{-} + t_{\alpha } \frac{S}{\sqrt{n} })$

$(2.92 - 2.093(\frac{0.16}{\sqrt{20} } ,2.92+2.093(\frac{0.16}{\sqrt{20} } )$

(2.92-0.0748,2.92+0.0748)

(2.84 ,2.99)

Therefore the 95% of confidence intervals

(2.84 ,2.99)

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