Question

test the claim at 5% significance.
round to 4 decimals when appropriate.
test statistic:
degrees of freedom:
p - value:
decision rule:
did something significant happen?
there enough evidence to conclude

Explanation:

Step1: Calculate the chi - square test statistic

The formula for the chi - square test statistic is \(\chi^{2}=\sum\frac{(O - E)^{2}}{E}\), where \(O\) is the observed frequency and \(E\) is the expected frequency.
For category A: \(\frac{(49 - 49)^{2}}{49}=0\)
For category B: \(\frac{(39 - 61.25)^{2}}{61.25}=\frac{(- 22.25)^{2}}{61.25}=\frac{495.0625}{61.25}=8.0825\)
For category C: \(\frac{(88 - 87.75)^{2}}{87.75}=\frac{(0.25)^{2}}{87.75}=\frac{0.0625}{87.75}\approx0.0007\)
For category D: \(\frac{(10 - 24.5)^{2}}{24.5}=\frac{(-14.5)^{2}}{24.5}=\frac{210.25}{24.5}=8.5816\)
For category F: \(\frac{(59 - 24.5)^{2}}{24.5}=\frac{(34.5)^{2}}{24.5}=\frac{1190.25}{24.5}=48.5816\)
Sum these values: \(0 + 8.0825+0.0007 + 8.5816+48.5816=65.2464\)

Step2: Determine the degrees of freedom

The formula for degrees of freedom in a chi - square goodness - of - fit test is \(df = k-1\), where \(k\) is the number of categories. Here \(k = 5\), so \(df=5 - 1=4\)

Step3: Find the p - value

Using a chi - square distribution table or a statistical software, for \(\chi^{2}=65.2464\) and \(df = 4\), the p - value is extremely small (close to \(0\))

Step4: Make a decision

Since the p - value (\(\approx0\)) is less than the significance level \(\alpha = 0.05\), we reject the null hypothesis.

Answer:

There is enough evidence to conclude that the teacher's distribution of grades is not as the teacher claims.