Question

Two independent random samples were selected from normally distributed populations to conduct an experiment on the variability of responses of two different experimental procedures that may or may not be the same. The sample sizes, means, and variances are shown in the following table

a. Test the null hypothesis\({H_0}:{\mu _d} = 0\)against\({H_0}:{\mu _d} \ne 0\), where,\({\mu _d} = {\mu _1} - {\mu _2}\). Compare\(\alpha = 0.05\)with the p-value of the test

Short answer

we fail to reject the null hypothesis.

Step-by-step solution

Given Information

The sample sizes are 14 and 16

The means are 44 and 41.

The variances are 569.2 and 810

The hypothesis are given by

\(\begin{aligned}{l}{H_0}:{\mu _d} = 0\\{H_a}:{\mu _d} \ne 0\end{aligned}\)

Test statistic

For

\(\begin{aligned}{l}{n_1} &= 14,\,\,{n_2} &= 16\\{{\bar x}_1} = 44,\,\;{{\bar x}_2} &= 41\\s_1^2 &= 569.2,\,s_2^2 &= 810\end{aligned}\)

The test statistic is computed as

\(\begin{aligned}{c}t &= \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - {D_0}}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}\\ &= \frac{{\left( {44 - 41} \right) - 0}}{{\sqrt {\frac{{569.2}}{{14}} + \frac{{810}}{{16}}} }}\\ &= \frac{3}{{\sqrt {40.65 + 50.62} }}\\ &= \frac{3}{{9.55}}\\ = 0.314\end{aligned}\)

Therefore, the test statistic is 0.314.

P-value

Degrees of freedom are calculated as

\(\begin{aligned}{c}df &= {n_1} + {n_2} - 2\\ &= 14 + 16 - 2\\ = 30 - 2\\ &= 28\end{aligned}\)

For \(df = 28,\alpha = 0.05\,and\,t = 0.314\)

The p-value is 0.755.

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