SOLUTION: The student in district high school is estimated to be76600. Among these student, 52% were in private schools while the rest were in public school. Among those in private, the fema

Question 1170491: The student in district high school is estimated to be76600. Among these student, 52% were in private schools while the rest were in public school. Among those in private, the female students occupied 55%, while in the public the percentage of the female was 44%. If one wants to make an investigation on the students performance in the district what is the sample size needed for both sexes and in both private and public institutions? Consider the following and determine sample size with proportion allocation. Level of confidence ( cl=95%) sampling error (e=5%) level of variability (=40%) response rate (Rr=70%)
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Let's break down this problem step by step to determine the necessary sample size using proportional allocation.
**1. Calculate the Number of Students in Private and Public Schools**
* Total students (N) = 76,600
* Private school students (Np) = 52% of 76,600 = 0.52 * 76,600 = 39,832
* Public school students (Nu) = 48% of 76,600 = 0.48 * 76,600 = 36,768
**2. Calculate the Number of Female and Male Students in Each School Type**
* **Private Schools:**
* Female students (Fp) = 55% of 39,832 = 0.55 * 39,832 = 21,907.6 (approximately 21,908)
* Male students (Mp) = 45% of 39,832 = 0.45 * 39,832 = 17,924.4 (approximately 17,924)
* **Public Schools:**
* Female students (Fu) = 44% of 36,768 = 0.44 * 36,768 = 16,177.92 (approximately 16,178)
* Male students (Mu) = 56% of 36,768 = 0.56 * 36,768 = 20,590.08 (approximately 20,590)
**3. Determine the Z-score for the Confidence Level**
* Confidence level (CL) = 95%
* Z-score for 95% confidence level = 1.96 (from the standard normal distribution table)
**4. Calculate the Sample Size for the Entire District**
We will use the formula for sample size calculation with a proportion:
n = (Z^2 * p * (1 - p)) / e^2
Where:
* n = sample size
* Z = Z-score (1.96)
* p = level of variability (0.40)
* e = sampling error (0.05)
n = (1.96^2 * 0.40 * 0.60) / 0.05^2
n = (3.8416 * 0.24) / 0.0025
n = 0.921984 / 0.0025
n = 368.7936 (approximately 369)
**5. Adjust for Response Rate**
We need to adjust the sample size for the response rate of 70% (0.70).
Adjusted sample size = n / response rate
Adjusted sample size = 369 / 0.70
Adjusted sample size = 527.14 (approximately 528)
**6. Allocate the Sample Size Proportionally**
We need to proportionally allocate the adjusted sample size (528) among the four groups:
* **Private Female (Fp):**
* Proportion = 21,908 / 76,600 ≈ 0.286
* Sample size = 0.286 * 528 ≈ 151
* **Private Male (Mp):**
* Proportion = 17,924 / 76,600 ≈ 0.234
* Sample size = 0.234 * 528 ≈ 123
* **Public Female (Fu):**
* Proportion = 16,178 / 76,600 ≈ 0.211
* Sample size = 0.211 * 528 ≈ 111
* **Public Male (Mu):**
* Proportion = 20,590 / 76,600 ≈ 0.269
* Sample size = 0.269 * 528 ≈ 142
**Results:**
* **Private Female:** 151 students
* **Private Male:** 123 students
* **Public Female:** 111 students
* **Public Male:** 142 students
**Summary**
To conduct the investigation with a 95% confidence level, 5% sampling error, 40% variability, and a 70% response rate, you would need a sample size of approximately 528 students, allocated as follows:
* 151 female students from private schools.
* 123 male students from private schools.
* 111 female students from public schools.
* 142 male students from public schools.

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