5. A professor teaching introductory psychology gave the grades that follow. According to the normal curve, she should have given about 4% A’s, 14% B’s, 64% C’s, 14% D’s, and 4% F’s. Do a Chi-Square to determine if the actual grade assignment significantly departed from the normal distribution. α=.05

A’s B’s C’s D’s F’s
25 35 60 30 20

X^2 = ∑ (O-E)^2/E, where O = observed frequency and E = expected frequency.

∑ = sum of all the cells.

E = (column total * row total)/grand total

df = n - 1, where n = number of cells

Look up value in X^2 table in the back of your textbook.

To determine if the grade assignment significantly departed from the normal distribution, we can use a Chi-Square test.

Step 1: Set up the null hypothesis (H0) and the alternative hypothesis (Ha):
H0: The grade assignment follows the normal distribution.
Ha: The grade assignment does not follow the normal distribution.

Step 2: Calculate the expected frequencies for each grade based on the normal distribution percentages given:
Since there are 170 total grades assigned (25 A's + 35 B's + 60 C's + 30 D's + 20 F's), we can calculate the expected frequencies as follows:
Expected frequency for A's = 0.04 * 170 = 6.8
Expected frequency for B's = 0.14 * 170 = 23.8
Expected frequency for C's = 0.64 * 170 = 108.8
Expected frequency for D's = 0.14 * 170 = 23.8
Expected frequency for F's = 0.04 * 170 = 6.8

Step 3: Calculate the Chi-Square test statistic:
The Chi-Square test statistic is calculated using the formula:
χ² = Σ [(O - E)² / E]
where O is the observed frequency and E is the expected frequency.

Using the observed frequencies provided:
(O - E)² / E for each grade category is:
(A): (25 - 6.8)² / 6.8 = 28.235
(B): (35 - 23.8)² / 23.8 = 1.817
(C): (60 - 108.8)² / 108.8 = 17.131
(D): (30 - 23.8)² / 23.8 = 0.925
(F): (20 - 6.8)² / 6.8 = 14.118

Adding up these values: 28.235 + 1.817 + 17.131 + 0.925 + 14.118 = 62.226

Step 4: Determine the degrees of freedom:
Degrees of freedom (df) is calculated as the number of categories minus 1. In this case, we have 5 categories, so df = 5 - 1 = 4.

Step 5: Look up the critical value in the Chi-Square distribution table:
With df = 4 and α = 0.05, we find the critical value to be 9.488.

Step 6: Compare the test statistic with the critical value:
Since 62.226 > 9.488, the test statistic is larger than the critical value, which means we reject the null hypothesis.

Step 7: Interpret the result:
Based on the Chi-Square test, there is enough evidence to suggest that the actual grade assignment significantly departed from the normal distribution in this class.