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tamaranim1 [39]

3 years ago

12

A student is running a 5-kilometer race. He runs 1 kilometer every 3 minutes. Select the function that describes his distance fr

om the finish line after x minutes. (1 point)
f(x) = − 1 over 3 x + 5

f(x) = − 1 over 5 x + 3

f(x) = 1 over 5 x + 5

f(x) = 1 over 3 x + 3

1 answer:

KonstantinChe [14]3 years ago

8 0

F(x)=1 over 5 x + 3 is the answer

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A recent college graduate is in the process of deciding which one of three graduate schools he should apply to. He decides to ju

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For this case after conduct the ANOVA procedure they got an statistic of:

$F = 8.61$

With a p value of :

$p_v =P(F_{2,15} > 8.61) = 0.003$

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$t = \frac{\bar X_i -\bar X_j}{s_p \sqrt{\frac{1}{n_i} +\frac{1}{n_j}}}$

For school 1 and 2 we have:

$t = \frac{646.7 -606.7}{52.54 \sqrt{\frac{1}{6} +\frac{1}{6}}}= 1.318$

For school 1 and 3

$t = \frac{646.7 -523.3}{52.54 \sqrt{\frac{1}{6} +\frac{1}{6}}}= 4.068$

For school 2 and 3 we have:

$t = \frac{606.7 -523.3}{52.54 \sqrt{\frac{1}{6} +\frac{1}{6}}}= 2.749$

So as we can see we have significant difference between the means of school 1 and 3, and school 2 and 3

Step-by-step explanation:

Previous concetps

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"

Solution to the problem

The hypothesis for this case are:

Null hypothesis: $\mu_{1}=\mu_{2}=\mu_{3}$

Alternative hypothesis: Not all the means are equal $\mu_{i}\neq \mu_{j}, i,j=1,2,3$

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

$SST=SS_{between}+SS_{within}$

For this case after conduct the ANOVA procedure they got an statistic of:

$F = 8.61$

With a p value of :

$p_v =P(F_{2,15} > 8.61) = 0.003$

Since the p value is lower than the significance level $\alpha=0.1$

We can reject the null hypothesis that the means are equal at the significance level provided.

For the other part we can calculate the 3 statistics with the following formula:

$t = \frac{\bar X_i -\bar X_j}{s_p \sqrt{\frac{1}{n_i} +\frac{1}{n_j}}}$

For school 1 and 2 we have:

$t = \frac{646.7 -606.7}{52.54 \sqrt{\frac{1}{6} +\frac{1}{6}}}= 1.318$

For school 1 and 3

$t = \frac{646.7 -523.3}{52.54 \sqrt{\frac{1}{6} +\frac{1}{6}}}= 4.068$

For school 2 and 3 we have:

$t = \frac{606.7 -523.3}{52.54 \sqrt{\frac{1}{6} +\frac{1}{6}}}= 2.749$

So as we can see we have significant difference between the means of school 1 and 3, and school 2 and 3

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3 years ago