In Exercises 5–20, assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of\({n_1} - 1\)and\({n_2} - 1\).)Blanking Out on Tests Many students have had the unpleasant experience of panicking on a test because the first question was exceptionally difficult. The arrangement of test items was studied for its effect on anxiety. The following scores are measures of “debilitating test anxiety,” which most of us call panic or blanking out (based on data from “Item Arrangement, Cognitive Entry Characteristics, Sex and Test Anxiety as Predictors of Achievement in Examination Performance,” by Klimko, Journal of Experimental Education, Vol. 52, No. 4.) Is there sufficient evidence to support the claim that the two populations of scores have different means? Is there sufficient evidence to support the claim that the arrangement of the test items has an effect on the score? Is the conclusion affected by whether the significance level is 0.05 or 0.01?

Questions Arranged from Easy to Difficult
24.64	39.29	16.32	32.83	28.02
33.31	20.60	21.13	26.69	28.9
26.43	24.23	7.10	32.86	21.06
28.89	28.71	31.73	30.02	21.96
25.49	38.81	27.85	30.29	30.72

Questions Arranged from Difficult to Easy
33.62	34.02	26.63	30.26
35.91	26.68	29.49	35.32
27.24	32.34	29.34	33.53
27.62	42.91	30.20	32.54

Data are given on the anxiety scores for the two different arrangements of questions on a test.

Null hypothesis:The two populations of scores have the same mean.

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

Alternative hypothesis:The two populations of scores have different means.

\({H_1}\):\({\mu _1} \ne {\mu _2}\).

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

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

Therefore, the mean score corresponding to the arrangement of questions from easy to difficult is equal to 27.12.

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

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

Therefore, the mean score corresponding to the arrangement of questions fromdifficult to easy is equal to 31.73.

The standard deviation ofthe scorescorresponding to the arrangement of questions from easy to difficult is equal to:

\(\begin{array}{c}{s_1} = \sqrt {\frac{{\sum\limits_{i = 1}^{{n_1}} {{{({x_i} - {{\bar x}_1})}^2}} }}{{{n_1} - 1}}} \\ = \sqrt {\frac{{{{\left( {24.64 - 27.12} \right)}^2} + {{\left( {39.29 - 27.12} \right)}^2} + .... + {{\left( {30.72 - 27.12} \right)}^2}}}{{25 - 1}}} \\ = 6.86\end{array}\)

Therefore, the standard deviation ofthe scorescorresponding to the arrangement of questions from easy to difficult is equal to 6.86.

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

\(\begin{array}{c}{s_2} = \sqrt {\frac{{\sum\limits_{i = 1}^{{n_2}} {{{({x_i} - {{\bar x}_2})}^2}} }}{{{n_2} - 1}}} \\ = \sqrt {\frac{{{{\left( {33.62 - 31.73} \right)}^2} + {{\left( {34.02 - 31.73} \right)}^2} + .... + {{\left( {32.54 - 31.73} \right)}^2}}}{{16 - 1}}} \\ = 4.26\end{array}\)

Therefore, the standard deviation of the scores corresponding to the arrangement of questions fromdifficult to easy is equal to 4.26.

Under null hypothesis,\({\mu _1} - {\mu _2} = 0\).

The test statistic is given by

\(\begin{array}{c}t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}\\ = \frac{{\left( {27.12 - 31.73} \right) - \left( 0 \right)}}{{\sqrt {\frac{{{{\left( {6.86} \right)}^2}}}{{25}} + \frac{{{{\left( {4.26} \right)}^2}}}{{16}}} }}\\ = - 2.657\end{array}\)

Thus, the calculated value of the test statistic is -2.657.

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

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

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

The value of the degrees of freedom is equal to the minimum of (24,15),which is 15.

Now see the t-distribution table for the two-tailed test at a 0.05 level of significance and for 15 degrees of freedom.

The critical values are -2.1314 and 2.1314. The p-value for the test statistic value equal to -2.657 is equal to 0.0180.

The value of the test statistic does not lie between the critical values, and the p-value is less than 0.05. Therefore, the null hypothesis is rejected.

There is enough evidence to support the claim that the arrangement of the questions has an effect on the score at a 0.05 significance level.

Let the level of significance be equal to 0.01.

Referring to the t-distribution table, the critical values of t at\(\alpha = 0.01\) with 15 degrees of freedom for a two-tailed test are -2.9467 and 2.9467.

The p-value remains the same and is equal to 0.0180.

The test statistic value equal to -2.657 lies between the two critical values, and the p-value is greater than 0.01.

There is not sufficient evidence to support the claim that the arrangement of the questions has an effect on the scoreat a 0.01 level of significance.

Thus, the conclusion of the claim changes with the change in thesignificance level.