A random sample of five pairs of observations were selected, one of each pair from a population with mean\({\mu _2}\), the other from a population with mean \({\mu _2}\).The data are shown in the accompanying table.

Pair	Value from population 1	Value from population 2
1	28	22
2	31	27
3	24	20
4	30	27
5	22	20

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.

The number of pairs, \({n_d} = 5\).

The data set is given as follows:

The Student’s t-test statistic for the small sample is,

\(t = \frac{{\bar d - {D_0}}}{{{s_d}/\sqrt {{n_d}} }}\)

Where \(\bar d\) is the sample mean difference, \({s_d}\) is the sample standard deviation difference, and \({n_d}\) is the number of pairs.

A random sample of five pairs of observations were selected, one of each pair from a population with\({\mu _1}\) and other with mean\({\mu _2}\).

The calculation table is given as follows:

The mean of the difference of the two population is computed as follows:

\(\begin{aligned}{c}\bar d &= \frac{{\sum {{d_i}} }}{{{n_d}}}\\ &= \frac{{19}}{5}\\ &= 3.8\end{aligned}\)

The standard deviation of the difference of the two population is computed as follows:

\(\begin{aligned}{c}{s_d} &= \sqrt {\frac{1}{{\left( {{n_d} - 1} \right)}}\left( {\sum {d_i^2} - \frac{{{{\left( {\sum {{d_i}} } \right)}^2}}}{{{n_d}}}} \right)} \\ &= \sqrt {\frac{1}{{\left( {5 - 1} \right)}}\left( {81 - \frac{{{{\left( {19} \right)}^2}}}{5}} \right)} \\ &= \sqrt {\frac{1}{4}\left( {81 - 72.2} \right)} \\ &= \sqrt {2.2} \\ &= 1.48\end{aligned}\)

Thus, the mean is 3.8 and the standard deviation is 1.48.

a.

The null and alternative hypothesis are:

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

The significance level is, \(\alpha = 0.05.\)

The test statistic is computed as:

\(\begin{aligned}{c}t &= \frac{{\bar d - {D_0}}}{{{s_d}/\sqrt {{n_d}} }}\\ &= \frac{{3.8 - 0}}{{1.48/\sqrt 5 }}\\ &= \frac{{3.8}}{{0.662}}\\ &= 5.7412\end{aligned}\)

This is a two-tailed test. The rejection region is, \(|t| > {t_{\frac{\alpha }{2}}}\).

The degree of freedom is computed as:

\(\begin{aligned}{c}\left( {n - 1} \right) &= \left( {5 - 1} \right)\\ &= 4\end{aligned}\)

The \({t_{\frac{\alpha }{2}}}\) value with \(\alpha = 0.05\) corresponding to the degree of freedom 4 obtained from the t-table is 2.776.

Here, \(|t| > {t_{\frac{\alpha }{2}}} \Rightarrow |5.7412| > 2.776.\) So, we reject the null hypothesis.

Hence, there is no sufficient evidence to claim that the difference of the mean pairs is zero.