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Anton [14]
3 years ago
12

Three assembly lines are used to produce a certain component for an airliner. To examine the production rate, a random

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
7 0

Answer:

a) Null hypothesis: \mu_A =\mu_B =\mu C

Alternative hypothesis: \mu_i \neq \mu_j, i,j=A,B,C

SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 =20.5  

SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 =12.333  

SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 =8.16667  

And we have this property  

SST=SS_{between}+SS_{within}  

The degrees of freedom for the numerator on this case is given by df_{num}=df_{within}=k-1=3-1=2 where k =3 represent the number of groups.

The degrees of freedom for the denominator on this case is given by df_{den}=df_{between}=N-K=3*6-3=15.

And the total degrees of freedom would be df=N-1=3*6 -1 =15

The mean squares between groups are given by:

MS_{between}= \frac{SS_{between}}{k-1}= \frac{12.333}{2}=6.166

And the mean squares within are:

MS_{within}= \frac{SS_{within}}{N-k}= \frac{8.1667}{15}=0.544

And the F statistic is given by:

F = \frac{MS_{betw}}{MS_{with}}= \frac{6.166}{0.544}= 11.326

And the p value is given by:

p_v= P(F_{2,15} >11.326) = 0.00105

So then since the p value is lower then the significance level we have enough evidence to reject the null hypothesis and we conclude that we have at least on mean different between the 3 groups.

b) (\bar X_B -\bar X_C) \pm t_{\alpha/2} \sqrt{\frac{s^2_B}{n_B} +\frac{s^2_C}{n_C}}

The degrees of freedom are given by:

df = n_B +n_C -2= 6+6-2=10

The confidence level is 99% so then \alpha=1-0.99=0.01 and \alpha/2 =0.005 and the critical value would be: t_{\alpha/2}=3.169

The confidence interval would be given by:

(43.333 -41.5) - 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}= 0.321

(43.333 -41.5) + 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}=3.345

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"

Part a  

Null hypothesis: \mu_A =\mu_B =\mu C

Alternative hypothesis: \mu_i \neq \mu_j, i,j=A,B,C

If we assume that we have 3 groups and on each group from j=1,\dots,6 we have 6 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 =20.5  

SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 =12.333  

SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 =8.16667  

And we have this property  

SST=SS_{between}+SS_{within}  

The degrees of freedom for the numerator on this case is given by df_{num}=df_{within}=k-1=3-1=2 where k =3 represent the number of groups.

The degrees of freedom for the denominator on this case is given by df_{den}=df_{between}=N-K=3*6-3=15.

And the total degrees of freedom would be df=N-1=3*6 -1 =15

The mean squares between groups are given by:

MS_{between}= \frac{SS_{between}}{k-1}= \frac{12.333}{2}=6.166

And the mean squares within are:

MS_{within}= \frac{SS_{within}}{N-k}= \frac{8.1667}{15}=0.544

And the F statistic is given by:

F = \frac{MS_{betw}}{MS_{with}}= \frac{6.166}{0.544}= 11.326

And the p value is given by:

p_v= P(F_{2,15} >11.326) = 0.00105

So then since the p value is lower then the significance level we have enough evidence to reject the null hypothesis and we conclude that we have at least on mean different between the 3 groups.

Part b

For this case the confidence interval for the difference woud be given by:

(\bar X_B -\bar X_C) \pm t_{\alpha/2} \sqrt{\frac{s^2_B}{n_B} +\frac{s^2_C}{n_C}}

The degrees of freedom are given by:

df = n_B +n_C -2= 6+6-2=10

The confidence level is 99% so then \alpha=1-0.99=0.01 and \alpha/2 =0.005 and the critical value would be: t_{\alpha/2}=3.169

The confidence interval would be given by:

(43.333 -41.5) - 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}= 0.321

(43.333 -41.5) + 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}=3.345

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3 years ago
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Answer:

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Step-by-step explanation:

Hi there,

In order to find the length of side EF, you should realize that this is an <u>isosceles triangle</u>. We know that two sides and two angles of this type of triangle are equal to each other. We also know that the lengths of the sides opposite of the congruent angles are equal to each other. Thus, we proved that length of side EF is equal to the length of side FG.

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m∠E + m∠G + m∠F = 180°

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m∠F = 134°

Hope this explanation helps you understand this problem. Cheers.

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3 years ago
Writing to explain what are compatible numbers? What compatible numbers would you pick to estimate the diferrent of 777-248? Why
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I hope I helped you!!!!<3
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