Pages were randomly selected by the author from The Bear and the Dragon by Tom Clancy, Harry Potter and the Sorcerer’s Stone by J. K. Rowling, and War and Peace by Leo Tolstoy. The Flesch Reading Ease scores for those pages are listed below. Do the authors appear to have the same level of readability?

Clancy	58.2	73.4	73.1	64.4	72.7	89.2	43.9	76.3	76.4	78.9	69.4	72.9
Rowling	85.3	84.3	79.5	82.5	80.2	84.6	79.2	70.9	78.6	86.2	74.0	83.7
Tolstoy	69.4	64.2	71.4	71.6	68.5	51.9	72.2	74.4	52.8	58.4	65.4	73.6

Data are given on the reading scores of three different samples of authors.

The level of significance is 0.05.

The null hypothesis to test the difference in the reading scores of the three samples is as

\({H_0}:{\mu _1} = {\mu _2} = {\mu _3}\),where\({\mu _i}\)is the readability score of each of the three authors.

There is no significant difference in the readability of the three authors.

The alternative hypothesis is as follows.

There is a significant difference in the readability of the three authors.

If the computed F-statistic is greater than the critical value, the null hypothesis is rejected at a 5% level of significance.

If the computed F-statistic is less than the critical value, the null hypothesis is failed to reject at a 5% level of significance.

Let n denote the sample sizes.

As the three samples are of equal size, n=12.

Let\({\bar x_i}\)denote the sample means.

The three sample means are computed as below.

\(\begin{array}{c}{{\bar x}_1} = \frac{{58.2 + 73.4 + .... + 72.9}}{{12}}\\ = 70.73\\{{\bar x}_2} = \frac{{85.3 + 84.3 + .... + 83.7}}{{12}}\\ = 80.75\\{{\bar x}_3} = \frac{{69.4 + 64.2 + .... + 73.6}}{{12}}\\ = 66.15\\\end{array}\)

Let\({s^2}_{\bar x}\)denote the variance of the sample means.

The mean of the three sample means is equal to

\(\begin{array}{c}\bar \bar x = \frac{{70.73 + 80.75 + 66.15}}{3}\\ = 72.54\end{array}\).

The variance of the sample means is computed as below.

\(\begin{array}{c}s_{\bar x}^2 = \frac{{\sum\limits_{i = 1}^3 {{{({{\bar x}_i} - \bar \bar x)}^2}} }}{{3 - 1}}\\ = \frac{{{{\left( {70.73 - 72.54} \right)}^2} + {{\left( {80.75 - 72.54} \right)}^2} + {{\left( {66.15 - 72.54} \right)}^2}}}{{3 - 1}}\\ = 55.75\end{array}\)

The variance between the sample means is equal to

\(\begin{array}{c}ns_{\bar x}^2 = 12\left( {55.75} \right)\\ = 669.001\end{array}\).

Therefore, the variance between the sample means is equal to 669.001.

Now, the three sample variances denoted by\({s_i}^2\)are computed as below.

\(\begin{array}{c}{s_1}^2 = \frac{{\sum\limits_{i = 1}^n {{{({x_{1i}} - {{\bar x}_1})}^2}} }}{{n - 1}}\\ = \frac{{{{\left( {58.2 - 70.73} \right)}^2} + {{\left( {73.4 - 70.73} \right)}^2} + .... + {{\left( {72.9 - 70.73} \right)}^2}}}{{12 - 1}}\\ = 128.28\end{array}\)

\(\begin{array}{c}{s_2}^2 = \frac{{\sum\limits_{i = 1}^n {{{({x_{2i}} - {{\bar x}_2})}^2}} }}{{n - 1}}\\ = \frac{{{{\left( {85.3 - 80.75} \right)}^2} + {{\left( {84.3 - 80.75} \right)}^2} + .... + {{\left( {83.7 - 80.75} \right)}^2}}}{{12 - 1}}\\ = 21.92\end{array}\)

\(\begin{array}{c}{s_3}^2 = \frac{{\sum\limits_{i = 1}^n {{{({x_{3i}} - {{\bar x}_3})}^2}} }}{{n - 1}}\\ = \frac{{{{\left( {69.4 - 66.15} \right)}^2} + {{\left( {64.2 - 66.15} \right)}^2} + .... + {{\left( {73.6 - 66.15} \right)}^2}}}{{12 - 1}}\\ = 61.75\end{array}\)

The variance within samples\(\left( {{s_p}^2} \right)\)is equal to

\(\begin{array}{c}{s_p}^2 = \frac{{{s_1}^2 + {s_2}^2 + {s_3}^2}}{3}\\ = \frac{{128.28 + 21.92 + 61.75}}{3}\\ = 70.648\end{array}\).

Therefore, the variance within samples is equal to 70.648.

Now, the F-statistic is computed as shown below.

\(\begin{array}{c}F = \frac{{{\rm{variance}}\;{\rm{between}}\;{\rm{samples}}}}{{{\rm{variance}}\;{\rm{within}}\;{\rm{samples}}}}\\ = \frac{{n{s_{\bar x}}^2}}{{{s_p}^2}}\\ = \frac{{669.001}}{{70.648}}\\ = 9.4695\end{array}\)

Therefore, the value of the F-statistic is equal to 9.4695.

Let k be the number of samples.

It is known that k=3 and n=12.

Thus, the degrees of freedom are calculated as shown below.

\(\begin{array}{c}df = \left( {k - 1,k\left( {n - 1} \right)} \right)\\ = \left( {3 - 1,3\left( {12 - 1} \right)} \right)\\ = \left( {2,33} \right)\end{array}\)

The critical value of F at a 5% level of significance with (2,33) degrees of freedom from the F-distribution table is equal to 3.2849.

It can be observed that

\(\begin{array}{c}F > {F_{crit}}\\\left( {9.4695} \right) > \left( {3.2849} \right)\end{array}\).

As the value of the F-statistic is greater than the critical value, the null hypothesis is rejected at a 5% level of significance.

Therefore, there is enough evidence to conclude that there is a significant difference in the readability of the authors.

In other words, the three authors do not appear to have the same level of readability.