The percentages of scores falling below the following z-score values for a particular raw score distribution are:
$\begin{array}{ll}
z \text {-score of }-2.04 & 5 \% \
z \text {-score of }-1.04 & 20 \% \
\text { z-score of 0 } & 55 \% \
\text { z-score of } 1.04 & 90 \% \
\text { z-score of } 2.04 & 100 \%
\end{array}$ Does this distribution differ significantly from a normal distribution? (Use ? = .05.)
A) No, statistic = 11.8
B) No, statistic = 6.7
C) Yes, statistic = 11.8
D) Yes, statistic = 6.7

Question

Quizplus · Accepted Answer

We can test whether the distribution differs significantly from a normal distribution using the chi-square goodness-of-fit test. The steps are as follows:

1. Calculate the expected frequencies assuming a normal distribution, using the formula E = N(Φ(z) - Φ(z-1)), where N is the sample size and Φ(z) is the standard normal cumulative distribution function for the corresponding z-score. For example, for the z-score of -2.04, we have E = 1000(Φ(-2.04) - Φ(-3.04)) = 1000(0.0207 - 0.0035) = 17.2.

2. Calculate the chi-square statistic, which is the sum of (O - E)^2/E for all categories, where O is the observed frequency and E is the expected frequency. For example, for the z-score of -2.04, we have (5 - 17.2)^2/17.2 = 3.33.

3. Determine the degrees of freedom, which is the number of categories minus 1. In this case, there are 5 categories, so df = 4.

4. Look up the critical value of chi-square with df = 4 and α = 0.05 in the chi-square distribution table. The critical value is 9.488.

5. Compare the calculated chi-square statistic to the critical value. If the calculated chi-square value is greater than the critical value, reject the null hypothesis that the distribution does not differ significantly from a normal distribution.

Using this procedure, we find that the calculated chi-square statistic is 6.7, which is less than the critical value of 9.488. Therefore, we do not reject the null hypothesis and conclude that the distribution does not differ significantly from a normal distribution at the 0.05 level of significance. The best choice is B.

The Percentages of Scores Falling Below the Following Z-Score Values