Answer
Yes.
Work Step by Step
a. State the hypotheses and identify the claim.
$H_o:$ he did not deviate from his usual pitch count
$H_a:$ he deviated from his usual pitch count
b. Find the critical value(s).
$\alpha=0.05, n=4, df=3, \chi^2_c=7.815$
c. Compute the test value.
Observed values $56, 30, 20, 5$ total 111
Expected values $111\times0.62=68.82, 111\times0.18=19.98, 111\times0.17=18.87, 111\times0.03=3.33$
$\chi^2=\frac{(56-68.82)^2}{68.82}+\frac{(30-19.98)^2}{19.98}+\frac{(20-18.87)^2}{18.87}+\frac{(5-3.33)^2}{3.33}=8.32$
d. Make the decision.
$\chi^2>7.815$ we reject the null hypothesis.
e. Summarize the results.
There is sufficient evidence at $\alpha=$ 0.05 that he deviated from his usual pitch count.
