# The Boom

Topics: Statistical hypothesis testing, Chi-square distribution, Null hypothesis Pages: 2 (430 words) Published: February 20, 2013
This section of sample problems and solutions is a part of The Actuary's Free Study Guide for Exam 3L, authored by Mr. Stolyarov. This is Section 55 of the Study Guide. See an index of all sections by following the link in this paragraph.

Problem
Acme Toy Company prints baseball cards. The company claims that 30% of the cards are rookies, 60% veterans, and 10% are All-Stars. The cards are sold in packages of 100. Suppose a randomly-selected package of cards has 50 rookies, 45 veterans, and 5 All-Stars. Is this consistent with Acme's claim? Use a 0.05 level of significance. Solution

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below: * State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis. * Null hypothesis: The proportion of rookies, veterans, and All-Stars is 30%, 60% and 10%, respectively. * Alternative hypothesis: At least one of the proportions in the null hypothesis is false. * Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis. * Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value. DF = k - 1 = 3 - 1 = 2

(Ei) = n * pi
(E1) = 100 * 0.30 = 30
(E2) = 100 * 0.60 = 60
(E3) = 100 * 0.10 = 10

Χ2 = Σ [ (Oi - Ei)2 / Ei ]
Χ2 = [ (50 - 30)2 / 30 ] + [ (45 - 60)2 / 60 ] + [ (5 - 10)2 / 10 ] Χ2 = (400 / 30) + (225 / 60) + (25 / 10) = 13.33 + 3.75 + 2.50 = 19.58 where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, Ei is the expected...