No Variation in a SampleAn experiment was conducted to test the effects of alcohol. Researchers measured the breath alcohol levels for a treatment group of people who drank ethanol and another group given a placebo. The results are given below (based on data from “Effects of Alcohol Intoxication on Risk Taking, Strategy, and Error Rate in Visuomotor Performance,” by Streufert et al., Journal of Applied Psychology,Vol. 77, No. 4). Use a 0.05 significance level to test the claim that the two sample groups come from populations with the same mean.

Treatment Group: \({{\bf{n}}_{\bf{1}}}{\bf{ = 22,}}\;{{\bf{\bar x}}_{\bf{1}}}{\bf{ = 0}}{\bf{.0049,}}\;{{\bf{s}}_{\bf{1}}}{\bf{ = 0}}{\bf{.015}}\)

Placebo Group: \({{\bf{n}}_{\bf{2}}}{\bf{ = 22,}}\;{{\bf{\bar x}}_{\bf{2}}}{\bf{ = 0}}{\bf{.000,}}\;{{\bf{s}}_{\bf{2}}}{\bf{ = 0}}{\bf{.000}}\)

The breath alcohol levels for a treatment group of people who drank ethanol and another group of people who received placebo are recorded.

The values are given as follows:

The sample size corresponding to the treatment group\(\left( {{n_1}} \right)\)is equal to 22.

The sample size corresponding to the treatment group\(\left( {{n_2}} \right)\)is equal to 22.

The sample mean alcohol level for the treatment group\(\left( {{{\bar x}_1}} \right)\)is equal to 0.0049.

The sample mean alcohol level for the placebo group\(\left( {{{\bar x}_2}} \right)\)is equal to 0.000.

The sample standard deviation of the alcohol levels for the treatment group\(\left( {{s_1}} \right)\)is equal to 0.015.

The sample standard deviation of the alcohol levels for the placebo group \(\left( {{s_2}} \right)\) is equal to 0.000.

The following hypotheses are noted:

Null Hypothesis: The two population means are equal.

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

Alternative Hypothesis: The two population means are not equal.

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

The test statistic value is computed below:

\(\begin{array}{c}t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{{s_1}^2}}{{n{}_1}} + \frac{{{s_2}^2}}{{{n_2}}}} }}\\ = \frac{{0.0049 - 0.0000}}{{\sqrt {\frac{{{{\left( {0.015} \right)}^2}}}{{22}} + \frac{{{{\left( {0.000} \right)}^2}}}{{22}}} }}\\ = 1.532\end{array}\)

Thus, t = 1.532.

The value of the degrees of freedom is equal to the smaller of\(\left( {{n_1} - 1} \right)\)and\(\left( {{n_2} - 1} \right)\).

Thus,

\(\begin{array}{c}{n_1} - 1 = 22 - 1\\ = 21\end{array}\)

\(\begin{array}{c}{n_2} - 1 = 22 - 1\\ = 21\end{array}\)

Therefore, the value of the degrees of freedom is equal to 21.

Referring to the t-distribution table, use the column for 0.05 (Area in two Tail),and use the row with the value of 21 degrees of freedom to find the critical value.

Thus, the critical values are -2.0796 and 2.0796.

Referring to the t-distribution table, the two-tailed p-value of t equal to 1.532 is equal to 0.13302.

Since the test statistic value lies between the two critical values and the p-value is greater than 0.05, the null hypothesis is failed to reject.

There is not sufficient evidence to reject the claim that thetwo population means are equal.