Question

a six - sided die is rolled 120 times. fill in the expected frequency column. then, conduct a hypothesis test to determine if the die is fair. the data in table 11.34 are the result of the 120 rolls.

the following are the null and alternative hypotheses.
( h_0 ): the die is fair.
( h_a ): the die is not fair.
( \bigcirc ) a. true
( \bigcirc ) b. false

Explanation:

Step1: Calculate expected frequency

For a fair six - sided die, the probability of each face \(P(X = i)=\frac{1}{6}\), \(i = 1,2,\cdots,6\). Given \(n = 120\) trials, the expected frequency \(E=n\times P(X = i)\). So \(E=120\times\frac{1}{6}=20\) for each face value.

Step2: Conduct hypothesis test (Chi - Square goodness - of - fit test)

The test statistic is \(\chi^{2}=\sum\frac{(O - E)^{2}}{E}\), where \(O\) is the observed frequency.
For face value \(1\): \(\frac{(15 - 20)^{2}}{20}=\frac{25}{20}=1.25\)
For face value \(2\): \(\frac{(29 - 20)^{2}}{20}=\frac{81}{20}=4.05\)
For face value \(3\): \(\frac{(16 - 20)^{2}}{20}=\frac{16}{20}=0.8\)
For face value \(4\): \(\frac{(15 - 20)^{2}}{20}=\frac{25}{20}=1.25\)
For face value \(5\): \(\frac{(30 - 20)^{2}}{20}=\frac{100}{20}=5\)
For face value \(6\): \(\frac{(15 - 20)^{2}}{20}=\frac{25}{20}=1.25\)

\(\chi^{2}=1.25 + 4.05+0.8 + 1.25+5+1.25=13.65\)

The degrees of freedom \(df=k - 1\), where \(k = 6\) (number of categories), so \(df=6 - 1 = 5\)

Using a significance level \(\alpha = 0.05\), the critical value \(\chi_{0.05,5}^{2}=11.070\)

Since \(\chi^{2}=13.65>\chi_{0.05,5}^{2}=11.070\), we reject the null hypothesis \(H_{0}\)

Answer:

b. FALSE