Questions: The call centre at the Austin Police Department receives hundreds of calls for service from citizens every week. To determine whether the phone calls are uniformly distributed among the days in the week, a random sample of 147 calls are studied for their days of occurrence. The results are shown below. Day Sun Mon Tue Wed Thu Fri Sat Frequency 22 18 17 25 25 19 21 a. Complete the table below with the expected frequencies, on the assumption that each day should be equally proportionally in calls received. Day Sun Mou Tue Wed Thu Fir Sat Expected frequency b. A χ^2 goodness of fit test is performed at a 5% significance level. i. State the null and alternative hypotheses. ii. Write down the degrees of freedom. iii. Using your graphics display calculator, find the χ^2 statistic and p-value. c. The critical value is 12.592. Determine the conclusion for the test, providing a reason.

The call centre at the Austin Police Department receives hundreds of calls for service from citizens every week. To determine whether the phone calls are uniformly distributed among the days in the week, a random sample of 147 calls are studied for their days of occurrence. The results are shown below.
Day Sun Mon Tue Wed Thu Fri Sat
Frequency 22 18 17 25 25 19 21
a. Complete the table below with the expected frequencies, on the assumption that each day should be equally proportionally in calls received.
Day Sun Mou Tue Wed Thu Fir Sat
Expected frequency
b. A χ^2 goodness of fit test is performed at a 5% significance level.
i. State the null and alternative hypotheses.
ii. Write down the degrees of freedom.
iii. Using your graphics display calculator, find the χ^2 statistic and p-value.
c. The critical value is 12.592. Determine the conclusion for the test, providing a reason.

Transcript text: The call centre at the Austin Police Department receives hundreds of calls for service from citizens every week. To determine whether the phone calls are uniformly distributed among the days in the week, a random sample of 147 calls are studied for their days of occurrence. The results are shown below. Day & Sun & Mon & Tue & Wed & Thu & Fri & Sat Frequency & 22 & 18 & 17 & 25 & 25 & 19 & 21 a. Complete the table below with the expected frequencies, on the assumption that each day should be equally proportionally in calls received. Day & Sun & Mou & Tue & Wed & Thu & Fir & Sat Expected frequency & & & & & & & b. A $\chi^{2}$ goodness of fit test is performed at a $5 \%$ significance level. i. State the null and alternative hypotheses. ii. Write down the degrees of freedom. iii. Using your graphics display calculator, find the $\chi^{2}$ statistic and $p$-value. c. The critical value is 12.592 . Determine the conclusion for the test, providing a reason.

Solution

Solution Steps

Step 1: Expected Frequencies

To determine if the phone calls are uniformly distributed among the days of the week, we first calculate the expected frequencies. Given that there are 147 calls and 7 days in a week, the expected frequency for each day is:

\[ E_i = \frac{147}{7} = 21.0 \]

Thus, the expected frequencies are:

\[ \text{Expected Frequencies} = [21.0, 21.0, 21.0, 21.0, 21.0, 21.0, 21.0] \]

Step 2: Hypotheses

We state the null and alternative hypotheses as follows:

Null Hypothesis ($H_0$): The phone calls are uniformly distributed among the days of the week.
Alternative Hypothesis ($H_a$): The phone calls are not uniformly distributed among the days of the week.

Step 3: Degrees of Freedom

The degrees of freedom for this test is calculated as:

\[ \text{Degrees of Freedom} = k - 1 = 7 - 1 = 6 \]

where $k$ is the number of categories (days of the week).

Step 4: Chi-Square Test Statistic and p-value

The Chi-Square test statistic is calculated as:

\[ \chi^2 = \sum_i \frac{(O_i - E_i)^2}{E_i} = 2.9524 \]

The corresponding p-value for this statistic is:

\[ P = P(\chi^2 > 2.9524) = 0.8148 \]

Step 5: Critical Value and Conclusion

The critical value for a Chi-Square test with 6 degrees of freedom at a significance level of $\alpha = 0.05$ is:

\[ \chi^2(0.95, 6) = 12.5916 \]

Final Answer

Since the calculated Chi-Square statistic $2.9524$ is less than the critical value $12.5916$ and the p-value $0.8148$ is greater than $\alpha = 0.05$, we fail to reject the null hypothesis.

Thus, we conclude that there is not enough evidence to suggest that the phone calls are not uniformly distributed among the days of the week.

\[ \boxed{\text{Fail to reject } H_0} \]

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