Question

Levene-Brown-Forsythe Test Repeat Exercise 16 “Blanking Out on Tests” using the Levene-Brown-Forsythe test.

Short answer

There is insufficient evidence to support the claim that the two populations of scores have the same amount of variation.

Step-by-step solution

Given information

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.

The claim is to be tested using the Levene-Brown-Forsythe test.

Transform the sample values

Transformed sample 1:

Sort the samples in ascending order in the table shown below and obtain the median.

7.1	16.32	20.6	21.06	21.13	21.96	24.23	24.64	25.49	26.43
26.69	27.85	28.02	28.71	28.89	28.9	30.02	30.29	30.72	31.73
32.83	32.86	33.31	38.81	39.29

Since the number of observations in this sample is odd (25), the following formula is used to compute the median:

\(\begin{array}{c}{\rm{Median}} = {\left( {\frac{{{n_1} + 1}}{2}} \right)^{th}}obs\\ = {\left( {\frac{{25 + 1}}{2}} \right)^{th}}obs\\ = {13^{th}}obs\\ = 28.02\end{array}\)

Now subtract the median from each sample value and take the absolute value of the difference.

The following table shows the absolute deviations:

Sample values (x)	\(\left( {x - {\rm{median}}} \right)\)	\(\left| {x - {\rm{median}}} \right|\)
24.64	-3.38	3.38
33.31	5.29	5.29
26.43	-1.59	1.59
28.89	0.87	0.87
25.49	-2.53	2.53
39.29	11.27	11.27
20.6	-7.42	7.42
24.23	-3.79	3.79
28.71	0.69	0.69
38.81	10.79	10.79
16.32	-11.7	11.7
21.13	-6.89	6.89
7.1	-20.92	20.92
31.73	3.71	3.71
27.85	-0.17	0.17
32.83	4.81	4.81
26.69	-1.33	1.33
32.86	4.84	4.84
30.02	2	2
30.29	2.27	2.27
28.02	0	0
28.9	0.88	0.88
21.06	-6.96	6.96
21.96	-6.06	6.06
30.72	2.7	2.7

Transformed sample 2:

Sort the samples in ascending order in the table shown below and obtain the median.

26.63	26.68	27.24	27.62	29.34	29.49	30.2	30.26	32.34	32.54
33.53	33.62	34.02	35.32	35.91	42.91

Since the number of observations in this sample is even (16), the following formula is used to compute the median:

\(\begin{array}{c}{\rm{Media}}{{\rm{n}}_2} = \frac{{{{\left( {\frac{{{n_2}}}{2}} \right)}^{th}}obs + {{\left( {\frac{{{n_2}}}{2} + 1} \right)}^{th}}obs}}{2}\\ = \frac{{{{\left( {\frac{{16}}{2}} \right)}^{th}}obs + {{\left( {\frac{{16}}{2} + 1} \right)}^{th}}obs}}{2}\\ = \frac{{{8^{th}}obs + {9^{th}}obs}}{2}\\ = \frac{{30.26 + 32.34}}{2}\\ = 31.3\end{array}\)

Now subtract the median from each sample value and take the absolute value of the difference.

The following table shows the absolute deviations:

Sample values (x)	\(\left( {x - {\rm{median}}} \right)\)	\(\left| {x - {\rm{median}}} \right|\)
33.62	2.32	2.32
35.91	4.61	4.61
27.24	-4.06	4.06
27.62	-3.68	3.68
34.02	2.72	2.72
26.68	-4.62	4.62
32.34	1.04	1.04
42.91	11.61	11.61
26.63	-4.67	4.67
29.49	-1.81	1.81
29.34	-1.96	1.96
30.2	-1.1	1.1
30.26	-1.04	1.04
35.32	4.02	4.02
33.53	2.23	2.23
32.54	1.24	1.24

Hypotheses

Let\({\mu _1}\)and\({\mu _2}\)be the population’smean of the transformed sample values corresponding to the arrangement of questions from easy to difficult and easy to difficult, respectively.

Null Hypothesis: The population’s mean of the transformed sample values corresponding to the arrangement of questions from easy to difficult is equal to the population’s mean of the transformed sample values corresponding to the arrangement of questions from difficult to easy.

Symbolically,

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

Alternate Hypothesis: The population’s mean of the transformed sample values corresponding to the arrangement of questions from easy to difficult is not equal to the population mean of the transformed sample values corresponding to the arrangement of questions from difficult to easy.

Symbolically,

\({H_0}:{\mu _1} \ne {\mu _2}\)

Compute the sample means and sample variances of both the transformed samples

Transformed sample 1:

Let the transformed sample values corresponding to the arrangement of question from easy to difficult is denoted by x.

The sample mean is computed as follows:

\(\begin{array}{c}{{\bar x}_1} = \frac{{\sum\limits_{i = 1}^{{n_1}} {{x_i}} }}{{{n_1}}}\\ = \frac{{3.38 + 5.29 + ..... + 2.7}}{{25}}\\ = 4.914\end{array}\)

The sample variance is computed as follows:

\(\begin{array}{c}s_1^2 = \frac{{\sum\limits_{i = 1}^{{n_1}} {{{({x_i} - \bar x)}^2}} }}{{{n_1} - 1}}\\ = \frac{{{{\left( {3.38 - 4.914} \right)}^2} + {{\left( {5.29 - 4.914} \right)}^2} + ...... + {{\left( {2.7 - 4.914} \right)}^2}}}{{25 - 1}}\\ = 4.766\end{array}\)

Transformed sample 2:

Let the transformed sample values corresponding to the arrangement of question from difficult to easy is denoted by x.

The sample mean is computed as follows:

\(\begin{array}{c}{{\bar x}_2} = \frac{{\sum\limits_{i = 1}^{{n_2}} {{x_i}} }}{{{n_2}}}\\ = \frac{{2.32 + 4.61 + ..... + 1.24}}{{16}}\\ = 3.296\end{array}\)

The sample variance is computed as follows:

\(\begin{array}{c}s_2^2 = \frac{{\sum\limits_{i = 1}^{{n_2}} {{{({x_i} - {{\bar x}_2})}^2}} }}{{{n_2} - 1}}\\ = \frac{{{{\left( {2.32 - 3.296} \right)}^2} + {{\left( {4.61 - 3.296} \right)}^2} + ...... + {{\left( {1.24 - 3.296} \right)}^2}}}{{16 - 1}}\\ = 6.759\end{array}\)

Compute the test statistic

Apply the t-test to compute the test statistic using the given formula.

\(t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}\).

Substitute the respective values in the above formula and simplify.

\(\begin{array}{c}t = \frac{{\left( {4.914 - 3.296} \right) - 0}}{{\sqrt {\frac{{22.89}}{{25}} + \frac{{6.759}}{{16}}} }}\\ = 1.403\end{array}\)

Thus, t is equal to 1.403.

State the critical values and the p-value

The degrees of freedom is the smaller of the two values:\(\left( {{n_1} - 1} \right)\)and\(\left( {{n_2} - 1} \right)\).

The values are computed below:

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

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

Thus, the degree of freedom is equal to 15.

The critical values can be obtained using the t distribution table with the degree of freedom equal to 15 and the significance level equal to 0.05 for a two-tailed test.

The critical values are -2.1314 and 2.1314.

The p-value for t equal to 1.403 is equal to 0.1810.

Interpretation

Since the test statistic value lies between the two values, the null hypothesis fails to be rejected.

Therefore, there is sufficient evidence to support the claim that the variance in the case of easy to difficult question arrangement is different from the variation in the case of difficult to easy question arrangement.