To calculate the 90% confidence interval for the proportion of individuals who are dissatisfied with their working conditions, let's go through the steps.

Step-by-Step Solution

1. Identify Given Information:

   • Sample size, $n = 175$
   • Number of dissatisfied individuals, $x = 59$
   • Confidence level = 90%

2. Calculate Sample Proportion ( $\hat{p}$ ): $\hat{p} = \frac{x}{n} = \frac{59}{175} = 0.3371$

3. Determine the Z-Score for 90% Confidence Level: For a 90% confidence level, the Z-score is approximately 1.645 (from the standard normal distribution).

4. Calculate the Standard Error (SE): $SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.3371 \times (1 - 0.3371)}{175}} = 0.0359$

5. Calculate the Margin of Error (ME): $ME = Z \times SE = 1.645 \times 0.0359 = 0.0590$

6. Find the Confidence Interval: $\text{Lower limit} = \hat{p} - ME = 0.3371 - 0.0590 = 0.2781$ $\text{Upper limit} = \hat{p} + ME = 0.3371 + 0.0590 = 0.3961$

Answer

• Lower limit: 0.28 (rounded to two decimal places)
• Upper limit: 0.40 (rounded to two decimal places)

Would you like additional details on any part of this calculation?

1. How do you interpret the meaning of a 90% confidence interval?
2. What changes if you increase the sample size?
3. How would a higher confidence level affect the interval width?
4. Why is it important to round intermediate values?
5. What assumptions are made when constructing a confidence interval for a proportion?

Tip: Confidence intervals provide a range of plausible values for the population parameter based on sample data and the chosen confidence level.