Question 1162467: An industrial engineer intends to use the mean of a random sample of size n = 150 to estimate the average mechanical aptitude (as measured by a certain test) of assembly line workers in a large industry. If on the b basis of experience, the engineer can assume that alpha = 6.2 for such data, what can be assert with probability 0.99 about the maximum size of his error?
Answer by CPhill(2261)   (Show Source):
You can put this solution on YOUR website! ```python?code_reference&code_event_index=2
import math
from scipy.stats import norm
# Given values
n = 150
sigma = 6.2
confidence_level = 0.99
# Calculate z-score for 0.99 probability (two-tailed)
# Alpha is 1 - confidence_level
alpha_val = 1 - confidence_level
z_score = norm.ppf(1 - alpha_val / 2)
# Calculate Margin of Error (E)
error = z_score * (sigma / math.sqrt(n))
print(f"{z_score=}")
print(f"{error=}")
```
```text?code_stdout&code_event_index=2
z_score=2.5758293035489004
error=1.3039566080286962
```
To determine the maximum size of the error for estimating the population mean with a specific probability, we use the formula for the **Margin of Error ($E$)**:
$$E = z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}$$
### 1. Identify the given information
* **Sample size ($n$):** 150
* **Population standard deviation ($\sigma$):** 6.2 (The "alpha" mentioned in the prompt refers to the population standard deviation for the data)
* **Confidence level (Probability):** 0.99
### 2. Find the critical value ($z$)
For a probability of 0.99, the significance level ($\alpha$) is $1 - 0.99 = 0.01$. Since we are estimating a range around the mean, we look for the $z$-score corresponding to $\alpha/2 = 0.005$ in the tails.
* The $z$-score for a 0.99 confidence level is approximately **2.576**.
### 3. Calculate the maximum error
Substitute the values into the formula:
$$E = 2.576 \cdot \frac{6.2}{\sqrt{150}}$$
$$E = 2.576 \cdot \frac{6.2}{12.2474}$$
$$E = 2.576 \cdot 0.5062$$
$$E \approx 1.304$$
### Final Assertion
With a probability of 0.99, the industrial engineer can assert that the maximum size of his error in estimating the average mechanical aptitude is **1.304**.
This means the engineer is 99% confident that the sample mean will not differ from the true population mean by more than **1.304 units**.
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