Chapter 13: Q. 13.48 (page 543)

Movie fans use the annual Leonard Maltin Movie Guide for facts, cast members, and reviews of more than $21000$ films. The movies are rated from $4$ stars $(4 )$ , indicating a very good movie, to 1 star $(1 )$ , which Leonard Maltin refers to as a BOMB. The following table gives the running times, in minutes, of a random sample of films listed in one year's guide.
At the $1 %$ significance level, do the data provide sufficient evidence to conclude that a difference exists in mean running times among films in the four rating groups? (Note: $T_{1} = 483, T_{2} = 573, T_{3} = 576$ , $T_{4} = 691$ , $Σ x_{1}^{2} = 232.117 .)$

Short Answer

Expert verified

Data do not provide sufficient evidence at the $1 %$ significance level since the p-value fails to reject the null hypothesis.

$P > 0.01 \Rightarrow F a i l t o R e j e c t H_{0}$ .

Step by step solution

Given information

The given table is

Explanation

The given data is

Find the SST, SSTR and SSE using the relation

$S S T = \sum x^{2} - \frac{{(\sum x)}^{2}}{n}$

$S S T = 232117 - \frac{(2323)^{2}}{24}$

$= 7.27 \times 10^{3}$

$S S T R = \frac{\sum {(x_{i})}^{2}}{n_{i}} - \frac{{(\sum x)}^{2}}{n}$

$SSTR = \frac{483^{2}}{6} + \frac{573^{2}}{6} + \frac{486^{2}}{6} + \frac{691^{2}}{6} - \frac{(2323)^{2}}{24}$

$= 3.63 \times 10^{3}$

$S S E = S S T - S S T R$

$= 3.64 \times 10^{3}$

Then,

$d f_{T} = k - 1$

$\Rightarrow 4 - 1 = 3$

$d f_{E} = n - k$

$\Rightarrow 24 - 4 = 20$

$M S T R = \frac{S S T R}{d f_{T}}$

$\Rightarrow \frac{3.63 \times 10^{3}}{3} = 1210$

$M S E = \frac{S S E}{d f_{E}}$

$\Rightarrow \frac{3.64 \times 10^{3}}{12} = 303.33$

$F = \frac{M S T R}{M S E}$

$\Rightarrow \frac{1210}{303.33} \approx 3.99$

Then put the ANOVA table

Data do not provide sufficient evidence at the $1 %$ significance level since the p-value fails to reject the null hypothesis.

$P > 0.01 \Rightarrow F a i l t o R e j e c t H_{0}$ .

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