document.write( "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? \n" ); document.write( "
Algebra.Com's Answer #854402 by CPhill(2261)\"\" \"About 
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```python?code_reference&code_event_index=2
\n" ); document.write( "import math
\n" ); document.write( "from scipy.stats import norm\r
\n" ); document.write( "\n" ); document.write( "# Given values
\n" ); document.write( "n = 150
\n" ); document.write( "sigma = 6.2
\n" ); document.write( "confidence_level = 0.99\r
\n" ); document.write( "\n" ); document.write( "# Calculate z-score for 0.99 probability (two-tailed)
\n" ); document.write( "# Alpha is 1 - confidence_level
\n" ); document.write( "alpha_val = 1 - confidence_level
\n" ); document.write( "z_score = norm.ppf(1 - alpha_val / 2)\r
\n" ); document.write( "\n" ); document.write( "# Calculate Margin of Error (E)
\n" ); document.write( "error = z_score * (sigma / math.sqrt(n))\r
\n" ); document.write( "\n" ); document.write( "print(f\"{z_score=}\")
\n" ); document.write( "print(f\"{error=}\")\r
\n" ); document.write( "\n" ); document.write( "```
\n" ); document.write( "```text?code_stdout&code_event_index=2
\n" ); document.write( "z_score=2.5758293035489004
\n" ); document.write( "error=1.3039566080286962\r
\n" ); document.write( "\n" ); document.write( "```
\n" ); document.write( "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 ($E$)**:\r
\n" ); document.write( "\n" ); document.write( "$$E = z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}$$\r
\n" ); document.write( "\n" ); document.write( "### 1. Identify the given information
\n" ); document.write( "* **Sample size ($n$):** 150
\n" ); document.write( "* **Population standard deviation ($\sigma$):** 6.2 (The \"alpha\" mentioned in the prompt refers to the population standard deviation for the data)
\n" ); document.write( "* **Confidence level (Probability):** 0.99\r
\n" ); document.write( "\n" ); document.write( "### 2. Find the critical value ($z$)
\n" ); document.write( "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.
\n" ); document.write( "* The $z$-score for a 0.99 confidence level is approximately **2.576**.\r
\n" ); document.write( "\n" ); document.write( "### 3. Calculate the maximum error
\n" ); document.write( "Substitute the values into the formula:
\n" ); document.write( "$$E = 2.576 \cdot \frac{6.2}{\sqrt{150}}$$
\n" ); document.write( "$$E = 2.576 \cdot \frac{6.2}{12.2474}$$
\n" ); document.write( "$$E = 2.576 \cdot 0.5062$$
\n" ); document.write( "$$E \approx 1.304$$\r
\n" ); document.write( "\n" ); document.write( "### Final Assertion
\n" ); document.write( "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**. \r
\n" ); document.write( "\n" ); document.write( "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**.
\n" ); document.write( "
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