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Anni [7]
3 years ago
14

Please answer this question based on the following data. SSTR = 6,750 H0: μ1= μ2=μ3=μ4 SSE = 8,000 Ha: at least one mean is diff

erent Nτ = 20 The null hypothesis is to be tested at the 5% level of significance. The p-value is less than .01 between .01 and .025 between .025 and .05 between .05 and .10
Mathematics
1 answer:
Nookie1986 [14]3 years ago
6 0

Answer:

P_v =P(F_{3,16}>4.5)=0.0180

So on this case the best option for the answer would be:

between .01 and .025

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"

If we assume that we have 4 groups and on each group from j=1,\dots,5 we have 5 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=Treatment}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 =6750  

SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 =8000  

And we have this property  

SST=SS_{between}+SS_{within}=6750+8000=14750  

The degrees of freedom for the numerator on this case is given by df_{num}=k-1=4-1=3 where k =4 represent the number of groups.

The degrees of freedom for the denominator on this case is given by df_{den}=df_{between}=N-K=20-4=16.

And the total degrees of freedom would be df=N-1=20 -1 =19

We can find the MSR=\frac{6750}{3}=2250

And MSE=\frac{8000}{16}=500

And the we can find the F statistic F=\frac{MSR}{MSE}=\frac{2250}{500}=4.5

And with that we can find the p value. On this case the correct answer would be 3 for the numerator and 16 for the denominator.  

P_v =P(F_{3,16}>4.5)=0.0180

So on this case the best option for the answer would be:

between .01 and .025

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Solve for x in the equation 2x^2+3x-7=x^2+5x+39
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Hey there, hope I can help!

\mathrm{Subtract\:}x^2+5x+39\mathrm{\:from\:both\:sides}
2x^2+3x-7-\left(x^2+5x+39\right)=x^2+5x+39-\left(x^2+5x+39\right)

Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
x^2-2x-46=0

Lets use the quadratic formula now
\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}
x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:} a=1,\:b=-2,\:c=-46: x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\left(-46\right)}}{2\cdot \:1}

\frac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \  \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \  \frac{2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}

Multiply the numbers 2 * 1 = 2
\frac{2+\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}

2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \  \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}

\mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \  \sqrt{\left(-2\right)^2+1\cdot \:4\cdot \:46} \ \textgreater \  \left(-2\right)^2=2^2, 2^2 = 4

\mathrm{Multiply\:the\:numbers:}\:4\cdot \:1\cdot \:46=184 \ \textgreater \  \sqrt{4+184} \ \textgreater \  \sqrt{188} \ \textgreater \  2 + \sqrt{188}
\frac{2+\sqrt{188}}{2} \ \textgreater \  Prime\;factorize\;188 \ \textgreater \  2^2\cdot \:47 \ \textgreater \  \sqrt{2^2\cdot \:47}

\mathrm{Apply\:radical\:rule}: \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} \ \textgreater \  \sqrt{47}\sqrt{2^2}

\mathrm{Apply\:radical\:rule}: \sqrt[n]{a^n}=a \ \textgreater \  \sqrt{2^2}=2 \ \textgreater \  2\sqrt{47} \ \textgreater \  \frac{2+2\sqrt{47}}{2}

Factor\;2+2\sqrt{47} \ \textgreater \  Rewrite\;as\;1\cdot \:2+2\sqrt{47}
\mathrm{Factor\:out\:common\:term\:}2 \ \textgreater \  2\left(1+\sqrt{47}\right) \ \textgreater \  \frac{2\left(1+\sqrt{47}\right)}{2}

\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1 \ \textgreater \  1+\sqrt{47}

Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.

\frac{-\left(-2\right)-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \  \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \  \frac{2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}

\frac{2-\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}

2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \  2-\sqrt{188} \ \textgreater \  \frac{2-\sqrt{188}}{2}

\sqrt{188} = 2\sqrt{47} \ \textgreater \  \frac{2-2\sqrt{47}}{2}

2-2\sqrt{47} \ \textgreater \  2\left(1-\sqrt{47}\right) \ \textgreater \  \frac{2\left(1-\sqrt{47}\right)}{2} \ \textgreater \  1-\sqrt{47}

Therefore our final solutions are
x=1+\sqrt{47},\:x=1-\sqrt{47}

Hope this helps!
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