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The mean value, sample size and standard deviation values are given for the amount of information recalled by 2 samples of eyewitnesses after a non-stressful interrogation and after a stressful interrogation.

It is claimed that stress decreases the amount of information recalled by an eyewitness.

Corresponding to the given claim, the following hypotheses are set up:

Null Hypothesis: The amount of information recalled after a non-stressful interrogation is equal to the amount of information recalled after a stressful interrogation.

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

Alternative Hypothesis: The amount of information recalled after a stressful interrogation is less than the amount of information recalled after a non-stressful interrogation.

\({H_1}:{\mu _1} > {\mu _2}\)

The test is right-tailed.

Sample Means:

Let\({\bar x_1}\)denote the mean amount of information recalled after a non-stressful interrogation.

The value of\({\bar x_1}\)is equal to 53.3.

Let\({\bar x_2}\)denote the mean amount of information recalled after a stressful interrogation.

The value of\({\bar x_2}\)is equal to 45.3.

Sample Sizes:

Let\({n_1}\)denote the sample size corresponding to the non-stressful interrogation.

The value of\({n_1}\)is equal to 40.

Let\({n_2}\)denote the sample size corresponding to the stressful interrogation.

The value of\({n_2}\)is equal to 40.

Pooled Variance:

Let\({s_1}\)denote the standard deviation of the amount of information recalled after a non-stressful interrogation.

The value of\({s_1}\)is equal to 11.6.

Let\({s_2}\)denote the standard deviation of the amount of information recalled after a stressful interrogation.

The value of\({s_2}\)is equal to 13.2.

The value of the pooled variance is computed below:

\(\begin{aligned} s_p^2 &= \frac{{\left( {{n_1} - 1} \right)s_1^2 + \left( {{n_2} - 1} \right)s_2^2}}{{\left( {{n_1} - 1} \right) + \left( {{n_2} - 1} \right)}}\\ &= \frac{{\left( {40 - 1} \right){{\left( {11.6} \right)}^2} + \left( {40 - 1} \right){{\left( {13.2} \right)}^2}}}{{\left( {40 - 1} \right) + \left( {40 - 1} \right)}}\\ &= 154.4\end{aligned}\)

The test statistic value is computed as follows:

\(\begin{aligned} t &= \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_p^2}}{{{n_1}}} + \frac{{s_p^2}}{{{n_2}}}} }}\;\;\;{\rm{where}}\;{\mu _1} - {\mu _2}\;is\;0\\ &= \frac{{\left( {53.3 - 45.3} \right) - 0}}{{\sqrt {\frac{{154.4}}{{40}} + \frac{{154.4}}{{40}}} }}\\ &= 2.879\end{aligned}\)

The value of the degrees of freedom is equal to:

\(\begin{aligned} df &= {n_1} + {n_2} - 2\\ &= 40 + 40 - 2\\ &= 78\end{aligned}\)

Referring to t-table:

The critical value of t for \(\alpha = 0.01\) and 78 degrees of freedom for a right-tailed test is equal to 2.3751.

The corresponding p-value is equal to 0.0026.

Since the test statistic (2.879) is greater than the critical value and the p-value is less than 0.01, the null hypothesis is rejected.

There is enough evidence to conclude thatstress decreases the amount of information recalled by an eyewitness.