Question 1198869: Calculate the standard error. May normality be assumed? (Round your answers to 4 decimal places.)
n = 27, m = 0.24
(b) n = 56, m = 0.43
(c)n = 123, m = 0.47
(d)n = 578, m = 0.004
Answer by textot(100)   (Show Source):
You can put this solution on YOUR website! **Formula for Standard Error of the Proportion**
The standard error of the proportion (SE) is given by:
SE = √[p * (1 - p) / n]
where:
* p is the sample proportion
* n is the sample size
**Normality Assumption**
We can generally assume normality of the sampling distribution of the proportion if the following conditions are met:
* **n * p ≥ 10**
* **n * (1 - p) ≥ 10**
**Now let's calculate the standard error for each case and check for normality:**
**a) n = 27, m = 0.24**
* SE = √[0.24 * (1 - 0.24) / 27] = √[0.24 * 0.76 / 27] ≈ 0.0758
* n * p = 27 * 0.24 = 6.48
* n * (1 - p) = 27 * 0.76 = 20.52
* **Normality may not be assumed** because n * p < 10.
**b) n = 56, m = 0.43**
* SE = √[0.43 * (1 - 0.43) / 56] = √[0.43 * 0.57 / 56] ≈ 0.0658
* n * p = 56 * 0.43 = 24.08
* n * (1 - p) = 56 * 0.57 = 31.92
* **Normality can be assumed** as both n * p and n * (1 - p) are greater than 10.
**c) n = 123, m = 0.47**
* SE = √[0.47 * (1 - 0.47) / 123] = √[0.47 * 0.53 / 123] ≈ 0.0448
* n * p = 123 * 0.47 = 57.81
* n * (1 - p) = 123 * 0.53 = 65.19
* **Normality can be assumed** as both n * p and n * (1 - p) are greater than 10.
**d) n = 578, m = 0.004**
* SE = √[0.004 * (1 - 0.004) / 578] = √[0.004 * 0.996 / 578] ≈ 0.0026
* n * p = 578 * 0.004 = 2.312
* n * (1 - p) = 578 * 0.996 = 575.688
* **Normality may not be assumed** because n * p < 10.
**Summary**
| Sample Size (n) | Sample Proportion (m) | Standard Error (SE) | Normality Assumed? |
|---|---|---|---|
| 27 | 0.24 | 0.0758 | No |
| 56 | 0.43 | 0.0658 | Yes |
| 123 | 0.47 | 0.0448 | Yes |
| 578 | 0.004 | 0.0026 | No |
I hope this helps! Let me know if you have any other questions.
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