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Two samples are considered.

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

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

a.

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

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

Subtract the sample values for the first sample for the sample mean. Take the absolute values of the deviations. Sort the absolute deviations in ascending order.

The following table shows the absolute deviations for the first sample:

The sorted values of the absolute deviations are shown below:

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

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

Subtract the sample values for the second sample for the sample mean. Take the absolute values of the deviations. Sort the absolute deviations in ascending order.

The following table shows the absolute deviations for the second sample:

The sorted values of the absolute deviations are shown below:

b.

The largest absolute value in the second sample is 11.182. Thus, theabsolute deviations in the first sample that are greater than 11.182 are 11.695, 12.175, and 20.015.

Therefore,\({c_1} = 3\)

The largest absolute value in the first sample is 20.015. Thus, none of the absolute deviations in the second sample are greater than 20.015.

Therefore, \({c_2} = 0\)

c.

The value of\({n_1}\)is equal to 25, and the value of\({n_2}\)is equal to 16.

Since\({n_1} \ne {n_2}\), the critical value is given by the formula mentioned below:

\({\rm{Critical}}\;V{\rm{alue}} = \frac{{\log \left( {\frac{\alpha }{2}} \right)}}{{\log \left( {\frac{{{n_1}}}{{{n_1} + {n_2}}}} \right)}}\)

Substitute\(\alpha = 0.05\),\({n_1} = 25\)and\({n_2} = 16\)in the above formula and simplify to compute the required value as follows:

\(\begin{array}{c}{\rm{Critical}}\;V{\rm{alue}} = \frac{{\log \left( {\frac{\alpha }{2}} \right)}}{{\log \left( {\frac{{{n_1}}}{{{n_1} + {n_2}}}} \right)}}\\ = \frac{{\log \left( {\frac{{0.05}}{2}} \right)}}{{\log \left( {\frac{{25}}{{25 + 16}}} \right)}}\\ = 7.4569\end{array}\)

d.

• If the value of\({c_1}\)is greater than the critical value, the conclusion of\(\sigma _1^2 > \sigma _2^2\)is made.
• If the value of\({c_2}\)is greater than the critical value, the conclusion of\(\sigma _2^2 > \sigma _1^2\)is made.
• If neither of the above conditions satisfy, the null hypothesis is failed to reject.

Here,

\(\begin{array}{c}{c_1} = 3 < 7.4568\\{c_2} = 0 < 7.4568\end{array}\)

Thus, the critical value is greaterthan both \({c_1}\)and\({c_2}\). This implies that 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.