Chi-Square Test - Is this right?

Question

Below is a problem that I did and my answer matched the back of the book, but I do not have confidence that I did it right. Could somebody confirm (or deny) that my solution is correct?

Bob

Problem:
Over a long period of time the grades given by a group of instructors in a particular course have averaged $12\% A's, 18\% B's, 40\% C's, 18\%D's$ and $12\%F's$. A new instructor gives $22 A's$, $34 B's$, $66 C's$, $16 D's$ and $12 F's$ during two semesters. Determine at a $0.05$ significance level whether the new instructor is following the grading pattern set by the others.
Answer:
We assign the grades the numbers, $4$, $3$, $2$, $1$ and $0$ for the grades $A$, $B$, $C$, $D$ and $F$, respectively. \begin{eqnarray*} n &=& 22 + 34 + 66 + 16 + 12 = 150 \\ \end{eqnarray*} The expected number of A's is $18$. The expected number of B's is $27$. The expected number of C's is $60$. The expected number of D's is $27$. The expected number of F's is $18$. \begin{eqnarray*} df &=& 5 - 1 = 4 \\ \chi^2 &=& \frac{(22-18)^2}{18} + \frac{(34-27)^2}{27} + \frac{(66-60)^2}{60} + \frac{(16-27)^2}{27} + \frac{(12-18)^2}{18} \\ \chi^2 &=& \frac{16}{18} + \frac{(7)^2}{27} + \frac{(6)^2}{60} + \frac{(11)^2}{27} + \frac{36}{18} \\ \chi^2 &=& \frac{8}{9} + \frac{49}{27} + \frac{36}{60} + \frac{121}{27} + 2 \\ \chi^2 &=& \frac{8}{9} + \frac{49}{27} + \frac{36}{60} + \frac{121}{27} + 2 \\ \chi^2 &=& 9.7851 \\ \end{eqnarray*} In this case we have $P(\chi^2 < 9.7851) = 0.96$, hence we reject the null hypothesis and conclude the new instructor grades are not following the pattern of the old instructors.

Answer 1

That's correct!

Let $$H_0 :p_1=0.12, p_2=0.18, p_3=0.4, p_4=0.18, p_5=0.12$$

$$H_a: \text{ at least one of these probabilities does not hold}$$

R statistical software confirms your result.

observed <- c(22,34,66,16,12)
expected <- c(.12,.18,.4,.18,.12)
chisq.test(x=observed,p=expected)

    Chi-squared test for given probabilities

data:  observed
X-squared = 9.7852, df = 4, p-value = 0.04421

Our p-value is $0.04421$ so at $\alpha=0.05$ we reject the null hypothesis and conclude that the new instructor is not following the grading pattern set by the others.

Note: Your method of finding $P(\chi^2 \lt 9.785)=0.96$ and concluding significance since $0.96\gt 0.95$ works but is unconventional. Better to just say $P(\chi^2 \gt 9.785)=0.04$