Independent random samples were selected from two normally distributed populations to test the hypothesis\({H_0}:{\mu _1} - {\mu _2} \ge 0\)against\({H_0}:{\mu _1} - {\mu _2} < 0\). Assume the two populations have equal variances. The sample sizes, means, and standard deviations are shown in the following table.

a. Test the above hypotheses at the 5% level of significance

The sample sizes are 10 and 15

The mean are 43.3 and 50.2

The standard deviations are 18.3 and 15.2

The hypothesis are given by

\(\begin{aligned}{l}{H_0}:{\mu _d} \ge 0\\{H_a}:{\mu _d} < 0\end{aligned}\)

For

\(\begin{aligned}{l}{n_1} &= 10,\,\,{n_2} &= 15\\{{\bar x}_1} &= 43.3,\,\;{{\bar x}_2} = 50.2\\{s_1} &= 18.3,\,{s_2} = 15.2\end{aligned}\)

The test statistic is computed as

\(\begin{aligned}{c}t &= \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - {D_0}}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}\\ &= \frac{{\left( {43.3 - 50.2} \right) - 0}}{{\sqrt {\frac{{334.89}}{{10}} + \frac{{231.04}}{{15}}} }}\\ &= \frac{{ - 6.9}}{{\sqrt {33.489 + 15.402} }}\\ &= \frac{{ - 6.9}}{{6.99}}\\ &= - 0.987\end{aligned}\)

Therefore, the test statistic is -0.987.

Degrees of freedom are calculated as

\(\begin{aligned}{c}df &= {n_1} + {n_2} - 2\\ &= 10 + 15 - 2\\ = 25 - 2\\ &= 23\end{aligned}\)

For \(df = 23,\alpha = 0.05\,\,and\,t = - 0.987\)

\(\)

The p-value is 0.1669.

Here, the p-value is greater than 0.05. Therefore, we fail to reject the null hypothesis.