Count Five Test for Comparing Variation in Two Populations Repeat Exercise 16 “Blanking Out on Tests,” but instead of using the F test, use the following procedure for the “count five” test of equal variations (which is not as complicated as it might appear).

a. For each value x in the first sample, find the absolute deviation \(\left| {x - \bar x} \right|\) , then sort the absolute deviation values. Do the same for the second sample.

b. Let \({c_1}\)be the count of the number of absolute deviation values in the first sample that is greater than the largest absolute deviation value in the second sample. Also, let \({c_2}\)be the count of the number of absolute deviation values in the second sample that are greater than the largest absolute deviation value in the first sample. (One of these counts will always be zero.)

c. If the sample sizes are equal (\({n_1} = {n_2}\)), use a critical value of 5. If\({n_1} \ne {n_2}\), calculate the critical value shown below.

\(\frac{{\log \left( {\frac{\alpha }{2}} \right)}}{{\log \left( {\frac{{{n_1}}}{{{n_1} + {n_2}}}} \right)}}\)

d. If \({c_1} \ge \) critical value, then conclude that \(\sigma _1^2 > \sigma _2^2\). If \({c_2} \ge \)critical value, then conclude that \(\sigma _2^2 > \sigma _1^2\). Otherwise, fail to reject the null hypothesis of \(\sigma _1^2 = \sigma _2^2\).

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.