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Two samples are considered.

One sample represents anxiety scores due to the arrangement of questions from easy to difficultin the test paper with a sample size equal to 25,and the other representsanxiety scores due to the arrangement of questions from difficult to easy in the test paper with a sample size equal to 16.

It is claimed that the variationin the scores corresponding to the arrangement of questions from easy to difficult is different from the variation in the scores corresponding to the arrangement of questions from difficult to easy.

Let\({\sigma _1}\)and\({\sigma _2}\)be the population standard deviationsof the scores corresponding to the arrangement of questions from easy to difficult and the arrangement from difficult to easy, respectively.

Null Hypothesis: The population standard deviation of the scores corresponding to the arrangement of questions from easy to difficult is equal to thepopulation standard deviation of the scores corresponding to the arrangement of questions from difficult to easy.

Symbolically,

\({H_0}:{\sigma _1} = {\sigma _2}\)

Alternate Hypothesis: The population standard deviation of the scores corresponding to the arrangement of questions from easy to difficult is not equal to thepopulation standard deviation of the scores corresponding to the arrangement of questions from difficult to easy.

Symbolically,

\({H_1}:{\sigma _1} \ne {\sigma _2}\)

The sample mean score corresponding to the arrangement of questions from easy to difficultis equal to:

\(\begin{array}{c}{{\bar x}_1} = \frac{{\sum\limits_{i = 1}^{{n_1}} {{x_i}} }}{{{n_1}}}\\ = \frac{{24.64 + 39.29 + .... + 30.72}}{{25}}\\ = 27.12\end{array}\)

The varianceofthe scorescorresponding to the arrangement of questions from easy to difficult is equal to:

\(\begin{array}{c}{s_{{\rm{easy}}\;{\rm{to}}\,{\rm{difficult}}}} = \frac{{\sum\limits_{i = 1}^{{n_1}} {{{({x_i} - {{\bar x}_1})}^2}} }}{{{n_1} - 1}}\\ = \frac{{{{\left( {24.64 - 27.12} \right)}^2} + {{\left( {39.29 - 27.12} \right)}^2} + .... + {{\left( {30.72 - 27.12} \right)}^2}}}{{25 - 1}}\\ = 47.02\end{array}\)

Therefore, the variance ofthe scorescorresponding to the arrangement of questions from easy to difficult is equal to 47.02.

The mean score corresponding to the arrangement of questions fromdifficult to easy is equal to:

\(\begin{array}{c}{{\bar x}_2} = \frac{{\sum\limits_{i = 1}^{{n_2}} {{x_i}} }}{{{n_2}}}\\ = \frac{{33.62 + 34.02 + .... + 32.54}}{{16}}\\ = 31.73\end{array}\)

The variance of the scorescorresponding to the arrangement of questions fromdifficult to easy is equal to:

\(\begin{array}{c}{s_{{\rm{difficult}}\;{\rm{to}}\;{\rm{easy}}}} = \frac{{\sum\limits_{i = 1}^{{n_2}} {{{({x_i} - {{\bar x}_2})}^2}} }}{{{n_2} - 1}}\\ = \frac{{{{\left( {33.62 - 31.73} \right)}^2} + {{\left( {34.02 - 31.73} \right)}^2} + .... + {{\left( {32.54 - 31.73} \right)}^2}}}{{16 - 1}}\\ = 18.15\end{array}\)

Therefore, the variance of the scores corresponding to the arrangement of questions fromdifficult to easy is equal to 18.15.

Since two independent samples involve a claim about the population standard deviation, apply an F-test.

Consider the larger sample variance to be\(s_1^2\)and the corresponding sample size to be\({n_1}\).

Here,\(s_1^2\)is the sample variance corresponding tothe arrangement of questions from easy to difficult is equal to 47.02.

\(s_2^2\)is the sample variance corresponding to thearrangement of questions fromdifficult to easy is equal to 18.15.

Substitute the respective values to calculate the F statistic:

\(\begin{array}{c}F = \frac{{s_1^2}}{{s_2^2}}\\ = \frac{{47.02}}{{18.15}}\\ = 2.591\end{array}\)

The value of the numerator degrees of freedom is equal to:

\(\begin{array}{c}{n_1} - 1 = 25 - 1\\ = 24\end{array}\)

The value of the denominator degrees of freedom is equal to:

\(\begin{array}{c}{n_2} - 1 = 16 - 1\\ = 15\end{array}\)

For the F test, the critical value corresponding to the right-tail is considered.

The critical value can be obtained using the F-distribution table with numerator degrees of freedom equal to25and denominator degrees of freedom equal to 15 for a right-tailed test.

The level of significance is equal to:

\(\begin{array}{c}\frac{\alpha }{2} = \frac{{0.05}}{2}\\ = 0.025\end{array}\)

Thus, the critical value is equal to 2.7006

The two-tailed p-value for F equal to 2.593 is equal to 0.0596.

Since the test statistic value is less than the critical value and the p-value is greater than 0.05, the null hypothesis is failed to reject.

Thus, there is not enough evidence to supportthe claimthat the two populations of scores have different amounts of variation.