# 7.4.1.5 - Example: Mean Quiz Score

7.4.1.5 - Example: Mean Quiz Score**Research question: **Is the mean quiz score different from 14 in the population?

StatKey was used to construct a randomization distribution:

#### Step 1: Check assumptions and write hypotheses

From the given StatKey output, the randomization distribution is approximately normal.

\(H_0\colon \mu = 14\)

\(H_a\colon \mu \ne 14\)

#### Step 2: Calculate the test statistic

\(test\;statistic=\dfrac{sample\;statistic-null\;parameter}{standard\;error}\)

The sample statistic is the mean in the original sample, 13.746 points. The null parameter is 14 points. And, the standard error, 0.142, can be found on the StatKey output.

\(test\;statistic=\dfrac{13.746-14}{0.142}=\dfrac{-0.254}{0.142}=-1.789\)

#### Step 3: Determine the p value

The p value will be the area on the z distribution that is more extreme than the test statistic of -1.789, in the direction of the alternative hypothesis:

This was a two-tailed test. The p value is the area in the left and right tails combined: \(p=0.0368074+0.0368074=0.0736148\)

#### Step 4: Make a decision

The p value (0.0736148) is greater than the standard 0.05 alpha level, therefore we fail to reject the null hypothesis.

#### Step 5: State a "real world" conclusion

There is not enough evidence to state that the mean quiz score in the population is different from 14 points.