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Given:

\(\begin{array}{l}{\rm{m = 18 }}\\{\rm{n = 18}}\end{array}\)

Let us assume:

\(\alpha = 0.05\)

Given claim: Difference

The null hypothesis states that there is no difference.

The alternative hypothesis states the given claim:

\(\begin{array}{l}{H_0}:{\mu _1} = {\mu _2}\\{H_a}:{\mu _1} \ne {\mu _2}\end{array}\)

Determine the rank of every data value. The smallest value receives the rank 1, the second smallest value receives the rank 2 , the third smallest value receives the rank 3 , and so on. If multiple data values have the same value, then their rank is the average of the corresponding ranks.

\({W_1}\)is the sum of the ranks in the smaller sample:

\(w = 402\)

Determine the mean and standard deviation:

\(\begin{array}{l}\\{\mu _R} = \frac{{{n_1}\left( {{n_1} + {n_2} + 1} \right)}}{2}\\{\mu _R} = \frac{{18(18 + 18 + 1)}}{2}\\{\mu _R} = 333\\{\sigma _R} = \sqrt {\frac{{{n_1}{n_2}\left( {{n_1} + {n_2} + 1} \right)}}{{12}}} \\{\sigma _R} = \sqrt {\frac{{18(18)(18 + 18 + 1)}}{{12}}} \\{\sigma _R} = 3\sqrt {111} \\\end{array}\)

The\(z\)-score is then the value of the test statistic decreased by the mean, divided by the standard deviation:

\(\begin{array}{l}z = \frac{{R - {\mu _R}}}{{{\sigma _R}}}\\z = \frac{{402 - 333}}{{3\sqrt {111} }}\\z \approx 2.18\end{array}\)

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true.
Determine the P-value using the normal probability table in the appendix:

\(\begin{array}{l}P = P(Z < - 2.18{\rm{ or }}Z > 2.18)\\P = 2P(Z < - 2.18)\\P = 2(0.0146)\\P = 0.0292\end{array}\)

If the probability (P-value) is less than the significant level , then reject the null hypothesis:

\(P < 0.05 \Rightarrow \)Reject\({H_0}\)

There is sufficient evidence to support the claim that the true average score depends on which learning method is used.