The superintendent of a large school district, having once had a course in probability and statistics, believes that the number of teachers absent on any given day has a Poisson distribution with parameter u. Use the accompanying data on absences for days to obtain a 95% large-sample CI for u. Number of absences 0 1 2 3 4567 89 10 Frequency 147 11 9 6 6 221 1 [Hint: The mean and variance of a Poisson variable both equal u, so X- u VH/n has approximately a standard normal distribution. Now proceed as in the derivation of the interval for p by making a probability statement (with probability 1 - a) and solving the resulting inequalities for u.] (Round your answers two decimal places.

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The superintendent of a large school district, having once had a course in probability and statistics, believes that the number of teachers absent on any given day has a Poisson distribution with parameter \( \mu \). Use the accompanying data on absences for 50 days to obtain a 95% large-sample confidence interval for \( \mu \).

| Number of absences | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|--------------------|---|---|---|---|---|---|---|---|---|---|----|
| Frequency          | 1 | 4 | 7 | 11| 9 | 6 | 6 | 2 | 2 | 1 | 1  |

*[Hint: The mean and variance of a Poisson variable both equal \( \mu \), so]*

\[ 
Z = \frac{\bar{X} - \mu}{\sqrt{\mu/n}} 
\]

has approximately a standard normal distribution. Now proceed as in the derivation of the interval for \( p \) by making a probability statement (with probability \( 1 - \alpha \)) and solving the resulting inequalities for \( \mu \). *(Round your answers to two decimal places.)*

\[ \left( \boxed{\phantom{XXX}}, \boxed{\phantom{XXX}} \right) \]

**Explanation of the Data:**

- The table shows the frequency of teacher absences over 50 days.
- The number of absences ranges from 0 to 10.
- Each number of absences has a corresponding frequency indicating how many days that number of absences occurred.

**Statistical Note:**

- The approach uses a Poisson distribution where both the mean and variance are equal to \( \mu \).
- The standard normal distribution is used to derive the confidence interval.
Transcribed Image Text:The superintendent of a large school district, having once had a course in probability and statistics, believes that the number of teachers absent on any given day has a Poisson distribution with parameter \( \mu \). Use the accompanying data on absences for 50 days to obtain a 95% large-sample confidence interval for \( \mu \). | Number of absences | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | |--------------------|---|---|---|---|---|---|---|---|---|---|----| | Frequency | 1 | 4 | 7 | 11| 9 | 6 | 6 | 2 | 2 | 1 | 1 | *[Hint: The mean and variance of a Poisson variable both equal \( \mu \), so]* \[ Z = \frac{\bar{X} - \mu}{\sqrt{\mu/n}} \] has approximately a standard normal distribution. Now proceed as in the derivation of the interval for \( p \) by making a probability statement (with probability \( 1 - \alpha \)) and solving the resulting inequalities for \( \mu \). *(Round your answers to two decimal places.)* \[ \left( \boxed{\phantom{XXX}}, \boxed{\phantom{XXX}} \right) \] **Explanation of the Data:** - The table shows the frequency of teacher absences over 50 days. - The number of absences ranges from 0 to 10. - Each number of absences has a corresponding frequency indicating how many days that number of absences occurred. **Statistical Note:** - The approach uses a Poisson distribution where both the mean and variance are equal to \( \mu \). - The standard normal distribution is used to derive the confidence interval.
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