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2 years ago

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Three runners, Dirk, Edith, and Foley all start at the same time for a $24$ km race, and each of them runs at a constant speed.

When Dirk finishes the race, Edith is $8$ km behind, and Foley is $12$ km behind. When Edith finishes the race, how far behind is Foley, in km

1 answer:

kompoz [17]2 years ago

4 0

Answer:

$12$ km

Step-by-step explanation:

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Solution :

We know that

$H_0: \mu_1 = \mu_2=\mu_3$

$H_1 :$ At least one mean is different form the others (claim)

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$\overline X_1 =50.5$ $\overline X_2 =50.07143$ $\overline X_3 =55.71429$

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The grand mean or the overall mean is(GM) :

$\overline X_{GM}=\frac{\sum \overline X}{N}$

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The variance between the groups

$s_B^2=\frac{\sum n_i\left( \overline X_i - \overline X_{GM}\right)^2}{k-1}$

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The Variance within the groups

$s_W^2=\frac{\sum(n_i-1)s_i^2}{\sum(n_i-1)}$

$=\frac{(14-1)19.96154+(14-1)14.68681+(7-1)36.57143}{(14-1)+(14-1)+(7-1)}$

$=\frac{669.8571}{32}$

= 20.93304

The F-test statistics value is :

$F=\frac{s_B^2}{s_W^2}$

$=\frac{63.55714}{20.93304}$

= 3.036212

Now since the 3.036 < 3.295, we do not reject the null hypothesis.

So there is no sufficient evidence to support the claim that there is a difference among the means.

The ANOVA table is :

Source Sum of squares d.f Mean square F

Between 127.1143 2 63.55714 3.036212

Within 669.8571 32 20.93304

Total 796.9714 34

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