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Question 1191336: he personnel director of a large corporation wishes to study absenteeism among clerical workers at the corporation's central office during the year. A random sample of 25 clerical workers reveals the following:
• Absenteeism: = 9.7 days, S = 4.0 days.
• 12 clerical workers were absent more than 10 days.
a. Construct a 95% confidence interval estimate for the mean number of absences for clerical workers during the year.
b. Construct a 95% confidence interval estimate for the population proportion of clerical workers absent more than 10 days during the year. Suppose that the personnel director also wishes to take a survey in a branch office. Answer these questions:
c. What sample size is needed to have 95% confidence in estimating the population mean absenteeism to within ±1.5 days if the population standard deviation is estimated to be 4.5 days?
d. How many clerical workers need to be selected to have 90% confidence in estimating the population proportion to within ± 0.075 if no previous estimate is available?
e. Based on (c) and (d), what sample size is needed if a single survey is being conducted?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem:
**a) 95% Confidence Interval for Mean Absences:**
Since the sample size is small (n=25) and the population standard deviation is unknown, we use the t-distribution.
1. **Find the critical t-value:** For a 95% confidence level and 24 degrees of freedom (n-1 = 25-1 = 24), the critical t-value is approximately 2.064 (you can find this using a t-table or calculator).
2. **Calculate the margin of error:**
Margin of Error = t * (S / √n) = 2.064 * (4.0 / √25) = 2.064 * 0.8 = 1.6512
3. **Construct the confidence interval:**
Confidence Interval = ȳ ± Margin of Error = 9.7 ± 1.6512
The 95% confidence interval is approximately (8.05, 11.35) days.
**b) 95% Confidence Interval for Population Proportion:**
1. **Calculate the sample proportion (p̂):**
p̂ = (Number of workers absent > 10 days) / (Total number of workers) = 12/25 = 0.48
2. **Find the critical z-value:** For a 95% confidence level, the critical z-value is 1.96.
3. **Calculate the margin of error:**
Margin of Error = z * √(p̂(1 - p̂) / n) = 1.96 * √(0.48 * 0.52 / 25) ≈ 0.196
4. **Construct the confidence interval:**
Confidence Interval = p̂ ± Margin of Error = 0.48 ± 0.196
The 95% confidence interval is approximately (0.284, 0.676).
**c) Sample Size for Mean Absences:**
1. **Use the sample size formula:**
n = (z * σ / E)²
Where:
* z is the critical z-value for 95% confidence (1.96)
* σ is the estimated population standard deviation (4.5 days)
* E is the desired margin of error (1.5 days)
2. **Calculate:**
n = (1.96 * 4.5 / 1.5)² = (5.88)² = 34.57
Since you can't have a fraction of a worker, round up to the nearest whole number. Therefore, a sample size of 35 workers is needed.
**d) Sample Size for Population Proportion:**
1. **Use the sample size formula (when no prior estimate is available):**
n = (z² * 0.25) / E²
Where:
* z is the critical z-value for 90% confidence (1.645)
* E is the desired margin of error (0.075)
2. **Calculate:**
n = (1.645² * 0.25) / 0.075² ≈ 120.278
Round up to the nearest whole number. A sample size of 121 workers is needed.
**e) Sample Size for Single Survey:**
Since a single survey is being conducted, you need to choose the larger of the two sample sizes calculated in parts (c) and (d). Therefore, a sample size of 121 workers is needed to satisfy both requirements.
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