A police department released the numbers of calls for the different days of the week during the month of​ October, as shown in the table to the right. Use a 0.010.01 significance level to test the claim that the different days of the week have the same frequencies of police calls. What is the fundamental error with this​ analysis?
Day nbspDay 
Sun
Mon
Tues
Wed
Thurs
Fri
Sat
 
FrequencyFrequency
157157
209209
227227
248248
173173
210210
230230
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Part 1
Determine the null and alternative hypotheses.
Upper H 0H0​: 
Police calls occur with the same frequency on the different days of the week.
Upper H 1H1​: 
At least one day has a different frequency of calls than the other days.
Part 2
Calculate the test​ statistic, chi squaredχ2.
chi squaredχ2equals= enter your response here ​(Round to three decimal places as​ needed.)

To test the claim that the different days of the week have the same frequencies of police calls, we will perform a chi-square goodness-of-fit test. Here's the step-by-step process:

### Part 1: Hypotheses

- **Null Hypothesis (H0):** Police calls occur with the same frequency on different days of the week.
- **Alternative Hypothesis (H1):** At least one day has a different frequency of calls than the other days.

### Part 2: Calculation of the Test Statistic, \( \chi^2 \)

1. **Observed Frequencies (O):**
   - Sun: 157
   - Mon: 209
   - Tues: 227
   - Wed: 248
   - Thurs: 173
   - Fri: 210
   - Sat: 230

2. **Expected Frequencies (E):**
   Since we are assuming that calls occur with the same frequency on different days of the week under the null hypothesis, we calculate the expected frequency by taking the average of the observed frequencies.

   \[
   E = \frac{\text{Total calls}}{\text{Number of days}} = \frac{157 + 209 + 227 + 248 + 173 + 210 + 230}{7} = \frac{1454}{7} = 207.714
   \]

3. **Chi-Square Test Statistic:**

   The formula for the chi-square test statistic is:

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

   where \( O_i \) are the observed frequencies and \( E_i \) are the expected frequencies.

   Calculating each component:

   \[
   \begin{aligned}
   \chi^2 &= \frac{(157 - 207.714)^2}{207.714} + \frac{(209 - 207.714)^2}{207.714} + \frac{(227 - 207.714)^2}{207.714} + \\
   & \quad \frac{(248 - 207.714)^2}{207.714} + \frac{(173 - 207.714)^2}{207.714} + \frac{(210 - 207.714)^2}{207.714} + \frac{(230 - 207.714)^2}{207.714} \\
   &= \frac{(-50.714)^2}{207.714} + \frac{(1.286)^2}{207.714} + \frac{(19.286)^2}{207.714} + \frac{(40.286)^2}{207.714} + \frac{(-34.714)^2}{207.714} + \frac{(2.286)^2}{207.714} + \frac{(22.286)^2}{207.714} \\
   &= \frac{2571.352}{207.714} + \frac{1.654}{207.714} + \frac{371.948}{207.714} + \frac{1622.968}{207.714} + \frac{1204.912}{207.714} + \frac{5.227}{207.714} + \frac{496.666}{207.714} \\
   &= 12.378 + 0.008 + 1.791 + 7.811 + 5.801 + 0.025 + 2.391 \\
   &= 30.205
   \end{aligned}
   \]

   So, the test statistic \( \chi^2 \approx 30.205 \) (rounded to three decimal places).

### Part 3: Critical Value and Conclusion

To complete the test, we compare the calculated chi-square statistic to the critical value from the chi-square distribution table with \( k-1 \) degrees of freedom (where \( k \) is the number of categories). Here, \( k = 7 \), so degrees of freedom \( = 7 - 1 = 6 \). At a significance level of \( 0.01 \), the critical value \( \chi^2_{0.01, 6} \) is approximately 16.812.

Since our test statistic \( 30.205 \) is greater than the critical value \( 16.812 \), we reject the null hypothesis.

### Conclusion
There is sufficient evidence at the \( 0.01 \) significance level to conclude that the frequencies of police calls differ across different days of the week.

### Fundamental Error with the Analysis
The fundamental error with this analysis could be assuming that the days of the week should naturally have the same frequencies without considering possible external factors (e.g., weekends might have different call patterns due to social behavior).

Would you like further details or have any questions?

Here are some related questions you might want to ask:
1. How do I interpret the chi-square test results?
2. What are the assumptions of the chi-square test?
3. Can we use this test for data with small sample sizes?
4. How do we calculate the degrees of freedom in chi-square tests?
5. What if the expected frequencies are very low?

**Tip:** When performing statistical tests, always check that the assumptions of the test are met to ensure valid results.

Understand and perform a chi-square test to analyze police calls frequencies across different days of the week. Step-by-step solution with calculations and conclusions at a 0.01 significance level.

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