Question

goodness of fit chi - squared hypothesis test ($\alpha = 0.05$) for the claim that all 6 outcomes of rolling 1 dice are equally likely. the number of trials is $n = 93$. enter the expected value for each possible outcome the table; round these expected values to four decimal places. round the following answers accurate to three decimal places. but retain unrounded numbers for future calculations. what is the chi - square test - statistic for this data? (report answer accurate to three decimal places.) $\chi^{2}=$ what is the p - value for this sample? (report answer accurate to 3 decimal places.) p - value = the p - value is... less than (or equal to) $\alpha$ greater than $\alpha$ this p - value leads to a decision to... reject the null hypothesis. fail to reject the null hypothesis. as such, the final conclusion is that... there is sufficient evidence to warrant rejection of the claim that all 6 categories are equally likely to be selected. there is not sufficient evidence to warrant rejection of the claim that all 6 categories are equally likely to be selected. question help: message instructor

Explanation:

Step1: Calculate expected frequency

When all 6 outcomes of rolling a die are equally likely and the number of trials $n = 93$, the expected frequency for each outcome is $E=\frac{n}{k}$, where $k = 6$ (number of possible outcomes). So $E=\frac{93}{6}=15.5000$.

Step2: Calculate chi - square test - statistic

The formula for the chi - square test - statistic is $\chi^{2}=\sum\frac{(O - E)^{2}}{E}$, where $O$ is the observed frequency and $E$ is the expected frequency.
For $O_1 = 14$: $\frac{(14 - 15.5)^{2}}{15.5}=\frac{(- 1.5)^{2}}{15.5}=\frac{2.25}{15.5}\approx0.145$
For $O_2 = 12$: $\frac{(12 - 15.5)^{2}}{15.5}=\frac{(-3.5)^{2}}{15.5}=\frac{12.25}{15.5}\approx0.790$
For $O_3 = 20$: $\frac{(20 - 15.5)^{2}}{15.5}=\frac{(4.5)^{2}}{15.5}=\frac{20.25}{15.5}\approx1.306$
For $O_4 = 16$: $\frac{(16 - 15.5)^{2}}{15.5}=\frac{(0.5)^{2}}{15.5}=\frac{0.25}{15.5}\approx0.016$
For $O_5 = 16$: $\frac{(16 - 15.5)^{2}}{15.5}=\frac{(0.5)^{2}}{15.5}=\frac{0.25}{15.5}\approx0.016$
For $O_6 = 15$: $\frac{(15 - 15.5)^{2}}{15.5}=\frac{(-0.5)^{2}}{15.5}=\frac{0.25}{15.5}\approx0.016$
$\chi^{2}=0.145 + 0.790+1.306 + 0.016+0.016+0.016=2.289$

Step3: Calculate degrees of freedom and p - value

The degrees of freedom $df=k - 1=6 - 1 = 5$. Using a chi - square distribution table or calculator, for $\chi^{2}=2.289$ and $df = 5$, the p - value is approximately $0.808$.

Step4: Make a decision

Since $\alpha=0.05$ and the p - value ($0.808$) is greater than $\alpha$, we fail to reject the null hypothesis.

Answer:

Expected Frequencies:
1: $15.5000$
2: $15.5000$
3: $15.5000$
4: $15.5000$
5: $15.5000$
6: $15.5000$
$\chi^{2}=2.289$
p - value $=0.808$
The p - value is Greater than $\alpha$
This P - Value leads to a decision to Fail to reject the null hypothesis.
As such, the final conclusion is that There is not sufficient evidence to warrant rejection of the claim that all 6 categories are equally likely to be selected.