SOLUTION: The 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 cle

Question 1179992: The 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: [25 Marks]
€ Absenteeism: X = 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 step-by-step:
**a. 95% Confidence Interval for Mean Absences:**
1. **Identify Given Information:**
* Sample mean (x̄) = 9.7 days
* Sample standard deviation (s) = 4.0 days
* Sample size (n) = 25
* Confidence level = 95%
2. **Determine Degrees of Freedom:**
* df = n - 1 = 25 - 1 = 24
3. **Find the Critical t-Value:**
* For a 95% confidence level and df = 24, the critical t-value (t*) is approximately 2.064 (from a t-table or calculator).
4. **Calculate the Margin of Error (E):**
* E = t* * (s / √n)
* E = 2.064 * (4.0 / √25)
* E = 2.064 * (4.0 / 5)
* E = 2.064 * 0.8 = 1.6512
5. **Construct the Confidence Interval:**
* Lower Bound = x̄ - E = 9.7 - 1.6512 = 8.0488
* Upper Bound = x̄ + E = 9.7 + 1.6512 = 11.3512
The 95% confidence interval is (8.0488, 11.3512) days.
**b. 95% Confidence Interval for Proportion of Absences > 10 Days:**
1. **Calculate Sample Proportion (p̂):**
* p̂ = 12 / 25 = 0.48
2. **Find the Critical z-Value:**
* For a 95% confidence level, the critical z-value (z*) is 1.96.
3. **Calculate the Margin of Error (E):**
* E = z* * √(p̂(1 - p̂) / n)
* E = 1.96 * √(0.48 * 0.52 / 25)
* E = 1.96 * √(0.2496 / 25)
* E = 1.96 * √0.009984
* E = 1.96 * 0.09992 ≈ 0.1958
4. **Construct the Confidence Interval:**
* Lower Bound = p̂ - E = 0.48 - 0.1958 = 0.2842
* Upper Bound = p̂ + E = 0.48 + 0.1958 = 0.6758
The 95% confidence interval is (0.2842, 0.6758).
**c. Sample Size for Mean Absenteeism (± 1.5 days):**
1. **Identify Given Information:**
* Margin of error (E) = 1.5 days
* Population standard deviation (σ) = 4.5 days
* Confidence level = 95% (z* = 1.96)
2. **Use the Sample Size Formula:**
* n = (z* * σ / E)²
* n = (1.96 * 4.5 / 1.5)²
* n = (8.82 / 1.5)²
* n = 5.88²
* n = 34.5744
3. **Round Up:**
* Since sample size must be a whole number, round up to 35.
**d. Sample Size for Proportion (± 0.075):**
1. **Identify Given Information:**
* Margin of error (E) = 0.075
* Confidence level = 90% (z* = 1.645)
* No previous estimate, so use p̂ = 0.5
2. **Use the Sample Size Formula:**
* n = (z*² * p̂ * (1 - p̂)) / E²
* n = (1.645² * 0.5 * 0.5) / 0.075²
* n = (2.706025 * 0.25) / 0.005625
* n = 0.67650625 / 0.005625
* n = 120.2678
3. **Round Up:**
* Round up to 121.
**e. Sample Size for a Single Survey:**
To conduct a single survey that satisfies both requirements, you need to use the *larger* of the two calculated sample sizes.
Therefore, the required sample size is 121.

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