15.34 In a four-digit lottery, each of the four digits is supposed to have the same probability of occurrence.
The table shows the frequency of occurrence of each digit for 89 consecutive daily fourdigit
drawings. Research question: At á = .01, can you reject the hypothesis that the digits are
from a uniform population? Why do the frequencies add to 356?
We do not have access to your table.
To answer the research question, we need to perform a statistical test. In this case, since we are comparing observed frequencies with expected frequencies, we can use the chi-square test for goodness of fit.
To conduct the chi-square test, we first need to state the null hypothesis (H0) and the alternative hypothesis (Ha). In this case, the null hypothesis is that the digits are from a uniform population, meaning each digit has an equal probability of occurrence. The alternative hypothesis is that the digits are not from a uniform population.
Next, we calculate the expected frequency for each digit under the assumption of a uniform population. Since there are four digits and each digit is supposed to have the same probability of occurrence, the expected frequency for each digit would be 356 / 4 = 89. The total frequency of 356 is expected because it represents the total number of observations over the 89 daily drawings.
Now, we can calculate the chi-square test statistic. The formula for the chi-square test statistic is:
χ² = ∑((Observed frequency - Expected frequency)² / Expected frequency)
Using the observed and expected frequencies from the table, we can calculate this test statistic.
The frequencies in the table add up to 356 because the table represents the distribution of digits over 89 consecutive daily drawings. Each drawing has four digits, so the total frequency is equal to 89 * 4 = 356.