columns (c = 4). The number of degrees of freedom for an r × c table is (r – 1)(c – 1). In this
question, (5 – 1)(4 – 1) = 4 × 3 = 12.
The correct answer is (c). In III, the expected count for each category in a goodness-of-fit test is
found by multiplying the proportion of the distribution of each category by the sample size. The
expected count for a test of independence is found by multiplying the row total by the column total
and then dividing by n .
Free-Response
1.
I. Let p 1 = the proportion of 0s, p 2 = the proportion of 1s, p 3 = the proportion of 2s, p 4 = the
proportion of 3s, and p 5 = the proportion of 4s.H 0 : p 1 = 0.24, p 2 = 0.41, p 3 = 0.27, p 4 = 0.07, p 5 = 0.01.H (^) A : Not all of the proportions stated in H 0 are correct.
II . We will use a chi-square goodness-of-fit test. The observed and expected values for the 500 trials
are shown in the following table:
We note that all expected values are at least 5, so the conditions necessary for the chi-square test
are present.
III . , df = 4 0.10 < P -value < 0.15 (from Table
C). Using the TI-83/84, χ 2 cdf(6.79,1000,40)= 0.147.
IV . The P- value is greater than any commonly accepted significance level. Hence, we do not reject H
0 and conclude that we do not have good evidence that the calculator is not correctly generating
values from B (4, 0.3).
For a 3 × 4 two-way table, df = (3 – 1)(4 – 1) = 6 0.005 < P -value < 0.01 (from Table C). The
finding is significant at the 0.01 level of significance. Using the TI-83/84, P -value = χ 2
cdf(17.2,1000,6)=0.009.
- (a) is not correct. For a given set of observations, they both do produce the same value of chi-square.
However, they differ in that they are different ways to design a study. (b) is correct. A test of