Question

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:$T_1=483,T_2=573,T_3=576$,$T_4=691$,$\Sigma x_1^2=232.117.)$

Short answer

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

$P>0.01\Rightarrow FailtoReject{H}_0$.

Step-by-step solution

Given information

The given table is

Explanation

The given data is

Find the SST, SSTR and SSE using the relation

$SST=\sum x^2-\frac{{\left(\sum x\right)}^2}{n}$

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

$=7.27\times {10}^3$

$SSTR=\frac{\sum {\left(x_i\right)}^2}{n_i}-\frac{{\left(\sum x\right)}^2}{n}$

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

$=3.63\times {10}^3$

$SSE=SST-SSTR$

$=3.64\times {10}^3$

Then,

$d{f}_T=k-1$

$\Rightarrow 4-1=3$

$d{f}_E=n-k$

$\Rightarrow 24-4=20$

$MSTR=\frac{SSTR}{d{f}_T}$

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

$MSE=\frac{SSE}{d{f}_E}$

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

$F=\frac{MSTR}{MSE}$

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

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.

$P>0.01\Rightarrow FailtoReject{H}_0$.