Question

(1 point) A random sample of 80 observations produced a mean of xˉ =35.5 from a population with a normal distribution and a standard deviation σ=2.61. (a) Find a 90% confidence interval for u ≤μ≤ (b) Find a 95% confidence interval for u ≤μ≤ (c) Find a 99% confidence interval for u ≤μ≤

Answer 1

Final Answers:

(a) \
`( \mu \)` lies between 34.98 and 36.02 with 90% confidence.

(b)
`\( \mu \)`lies between 34.87 and 36.13 with 95% confidence.

(c)
`\( \mu \)`lies between 34.67 and 36.33 with 99% confidence.

Explanation:

To find the confidence intervals, the
`formula`
`\( \bar{x} \pm z (\sigma)/(√(n)) \) is used, where \( \bar{x} \)` is the sample mean,
`\( \sigma \)` is the population standard deviation, n is the sample size, and z is the z-score associated with the desired confidence level.

For a 90% confidence level (a), the z-score is approximately 1.645. Plugging the values into the formula yields the interval [34.98, 36.02], indicating that we are 90% confident that the population mean falls within this range.

For a 95% confidence level (b), the z-score is about 1.96. Substituting the values gives the interval [34.87, 36.13], meaning we are 95% confident that
`\( \mu \)`lies within this interval.

Lastly, for a 99% confidence level (c), the z-score is approximately 2.576. Applying the formula provides the interval [34.67, 36.33], indicating that there is a 99% probability that
`\( \mu \)` falls within this range.

In summary, as the confidence level increases, the width of the interval increases as well, reflecting a higher certainty about the population mean
`\( \mu \)`within a broader range.