5. A random sample of 4500 U.S. employees was chosen and asked whether the "play hooky", that is, call in sick at least once a year when they simply need to relax; 1080 responded "yes". Use this information to set up a 95% confidence interval for the proportion, p, of all U.S. employees who play hooky. 1080 0.24 4500 4: 1-.24= .76 95% C.I 2c= O124 I 4500 1.96 %3D (0.275,0.2515)

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**Estimating the Proportion of U.S. Employees Who "Play Hooky"**

A random sample of 4500 U.S. employees was chosen, and each was asked whether they "play hooky" — that is, call in sick at least once a year when they simply need to relax. Out of these, 1080 employees responded “yes”. Using this data, we will set up a 95% confidence interval for the proportion, \( p \), of all U.S. employees who played hooky. 

### Step-by-Step Calculation:

1. **Sample Proportion (\(\hat{p}\))**: Calculate the sample proportion of employees who answered "yes".
   \[
   \hat{p} = \frac{x}{n} = \frac{1080}{4500} = 0.24
   \]

2. **Complement of the Sample Proportion (\(\hat{q}\))**: This is obtained by subtracting the sample proportion from 1.
   \[
   \hat{q} = 1 - \hat{p} = 1 - 0.24 = 0.76
   \]

3. **Critical Value (\(Z_c\))**: For a 95% confidence interval, the critical value is determined from the standard normal distribution (Z-table).
   \[
   Z_c = 1.96
   \]

4. **Standard Error (SE)**:
   \[
   SE = \sqrt{\frac{\hat{p} \cdot \hat{q}}{n}} = \sqrt{\frac{0.24 \cdot 0.76}{4500}}
   \]

5. **Margin of Error (ME)**: Multiply the critical value by the standard error.
   \[
   ME = Z_c \cdot SE = 1.96 \cdot \sqrt{\frac{0.24 \cdot 0.76}{4500}}
   \]

6. **Confidence Interval**: Add and subtract the margin of error from the sample proportion to obtain the confidence interval.
   \[
   \hat{p} \pm ME \implies 0.24 \pm 0.0125
   \]

7. The calculated interval is:
   \[
   (0.2275, 0.2525)
   \]

### Interpretation:
With 95% confidence, the
Transcribed Image Text:**Estimating the Proportion of U.S. Employees Who "Play Hooky"** A random sample of 4500 U.S. employees was chosen, and each was asked whether they "play hooky" — that is, call in sick at least once a year when they simply need to relax. Out of these, 1080 employees responded “yes”. Using this data, we will set up a 95% confidence interval for the proportion, \( p \), of all U.S. employees who played hooky. ### Step-by-Step Calculation: 1. **Sample Proportion (\(\hat{p}\))**: Calculate the sample proportion of employees who answered "yes". \[ \hat{p} = \frac{x}{n} = \frac{1080}{4500} = 0.24 \] 2. **Complement of the Sample Proportion (\(\hat{q}\))**: This is obtained by subtracting the sample proportion from 1. \[ \hat{q} = 1 - \hat{p} = 1 - 0.24 = 0.76 \] 3. **Critical Value (\(Z_c\))**: For a 95% confidence interval, the critical value is determined from the standard normal distribution (Z-table). \[ Z_c = 1.96 \] 4. **Standard Error (SE)**: \[ SE = \sqrt{\frac{\hat{p} \cdot \hat{q}}{n}} = \sqrt{\frac{0.24 \cdot 0.76}{4500}} \] 5. **Margin of Error (ME)**: Multiply the critical value by the standard error. \[ ME = Z_c \cdot SE = 1.96 \cdot \sqrt{\frac{0.24 \cdot 0.76}{4500}} \] 6. **Confidence Interval**: Add and subtract the margin of error from the sample proportion to obtain the confidence interval. \[ \hat{p} \pm ME \implies 0.24 \pm 0.0125 \] 7. The calculated interval is: \[ (0.2275, 0.2525) \] ### Interpretation: With 95% confidence, the
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