Question

a) Acme Toy Company prints baseball cards. The company claims that 30% of the cards are Rookies,
60% Veterans but not All-Stars, and 10% are Veteran All-Stars.
Suppose a random sample of 100 cards has 50 Rookies, 45 Veterans, and 5 Veterans All-Stars.
Table 1:
Rookies
Veterans
Veteran All-Star
Observed
50
45
5
Expected
30
60
10
i) Set-up a hypothesis test to test if this information is consistent with the Acme’s company claim? Use a
0.05 level of significance.

Answer 1

SOLUTION

Set the 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.

I will conduct a chi-square goodness of fit test of the null hypothesis.

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 and the determine the p-value.

$DF=k-1=3-1=2(E_i)=n*p_i$

$E_1=100*0.30=30$

$E_2=100*0.60=60$

$E_3=100*0.10=10$

$X^2=\sum [\frac{(O_i -E_i)^2}{E_I}]$

$X^2=[\frac{(50 -30)^2}{30}]+[\frac{(45 -60)^2}{60}]+[\frac{(5 -10)^2}{10}]$

$=\frac{400}{30}+\frac{225}{60}+\frac{25}{10}$

$=13.33+3.75+2.50$

$=19.58$

The P-value is the probability that a chi-square statistic having 2 degrees of freedom is more extreme than 19.58.

We use the Chi-Square Distribution Calculator to find $P(X^2\gt 19.58)=0.00005592=0.0001$

ANSWER: Since the P-value (0.0001) is less than the significance level (0.05), we reject the null hypothesis.