Question

An electronic parts factory produces resistors. Statistical analysis of the output suggests that resistances follow an approximately normal distribution with standard deviation of 0.6 ohms. A sample of 81 resistors has been collected, and the average resistance of the resistors x in the sample was measured. If we construct a 95% confidence interval for the population mean resistance m using x, what would the margin of error of estimating m by x be? Round your answer to 3 decimal places.

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Answer 1

The margin of error for a 95% confidence interval for the population mean resistance with a standard deviation of 0.6 ohms and a sample size of 81 is approximately 0.131 ohms.

Step-by-step explanation:

To calculate the margin of error for a 95% confidence interval for the population mean resistance, we use the z-score associated with the confidence level which, for 95%, is approximately 1.96. Given the standard deviation (σ) is 0.6 ohms and the sample size (n) is 81, the formula for the margin of error (EBM) is EBM = z * (σ / √n). Plugging in the values, we get EBM = 1.96 * (0.6 / √81), which simplifies to EBM = 1.96 * (0.6 / 9).

In this case, since the confidence level is 95%, the Z-score is 1.96. The sample standard deviation is 0.6 ohms, and the sample size is 81. Therefore, the margin of error is 1.96 * (0.6 / sqrt(81)), which is approximately 0.400 ohms.

Calculating that, we find EBM = 1.96 * 0.0667 ohms, which is approximately 0.131 ohms. Therefore, the margin of error for estimating the population mean resistance by the sample mean resistance is approximately 0.131 ohms when rounded to three decimal places.