Certified Quality Engineer (CQE) Practice Exam 2025 - Comprehensive All-in-One Study Guide for Exam Success!

Question: 1 / 400

For a sample with a mean length of 23.8 cm and an estimated standard deviation of 2.6 cm, what are the 95% confidence limits of the mean?

22.54 cm and 25.06 cm

22.83 cm and 24.77 cm

To determine the 95% confidence limits of the mean, we apply the formula for the confidence interval of the mean when the sample size is not specified, assuming we would use the t-distribution if the sample size were small but here treating it as normally distributed. The formula is given by:

Confidence Interval = Mean ± (Critical Value * Standard Deviation / √n)

In this case, since n (the sample size) is not provided, we typically assume a larger sample size for practical purposes to approximate using the normal distribution. The critical value for a 95% confidence interval from the z-table is approximately 1.96.

Using the provided mean of 23.8 cm and standard deviation of 2.6 cm, we can perform the calculations for the margin of error:

Margin of Error = Critical Value * (Standard Deviation / √n)

Assuming a sufficiently large n allows us to use the normal z-distribution:

- The margin of error can be approximated as:

Margin of Error ≈ 1.96 * 2.6

Calculating this gives:

- Margin of Error ≈ 5.096, or about 5.1 cm (for indicating the value).

Now

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22.99 cm and 24.61 cm

23.18 cm and 24.42 cm

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