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:

Randomization Test for a Mean Original Sample Generate 1 Sample Generate 10 Samples Generate 100 Samples Generate 1000 Samples 14 Reset Plot Edit Data Upload File Change Column(s) Show Data Table Custom Dataset Left Tail Two - Tail Right Tail 120 100 80 60 40 20 30 20 10 0 0 13.5 12.0 12.5 13.0 13.5 14.0 14.5 15.0 13.6 13.7 13.8 13.9 14.0 14.1 14.2 14.3 14.4 14.5 null = 14 13.746 n = 41, mean = 13.746 median = 13.98, stdev = 0.098

Step 1: Check assumptions and write hypotheses

From the given StatKey output, the sampling 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:

Distribution Plot Normal, Mean=0, StDev=1 0.0 0.1 0.2 0.3 0.4 0 X Density -1.789 0.0368074 0.0368074 1.78900

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 evidence that the mean quiz score in the population is different from 14 points.