130k views
1 vote
Assume that a sample is used to estimate a population mean. Find the 99.5% confidence interval for a sample of size 278 with a mean of 69.2 and a standard deviation of 19.1. Enter your answer as a tri-linear inequality accurate to 3 decimal places.

Confidence Interval: _______ < μ < _______

User OhMyGosh
by
7.8k points

1 Answer

5 votes

Final answer:

To find the 99.5% confidence interval, use the formula: Lower limit = sample mean - (critical value) * (standard deviation / square root of sample size), and Upper limit = sample mean + (critical value) * (standard deviation / square root of sample size). The critical value can be found by subtracting 0.5% from 100%, dividing it by 2, and looking it up in a standard normal distribution table. Plug the given values into the formula to calculate the confidence interval.

Step-by-step explanation:

To find the 99.5% confidence interval, we can use the formula:

Lower limit = sample mean - (critical value) * (standard deviation / square root of sample size)
Upper limit = sample mean + (critical value) * (standard deviation / square root of sample size)

First, we need to find the critical value.

Since we want a 99.5% confidence interval, we subtract 0.5% from 100% to get 99.5%. Then, we divide this value by 2 to get 0.25%. Looking up this value in a standard normal distribution table, we find that the critical value is approximately 2.807.

Now, we can substitute the given values into the formula to calculate the confidence interval:

Lower limit = 69.2 - 2.807 * (19.1 / sqrt(278))

Upper limit = 69.2 + 2.807 * (19.1 / sqrt(278))

Calculating these values gives:

Lower limit ≈ 67.921

Upper limit ≈ 70.479

Therefore, the 99.5% confidence interval for the population mean is approximately 67.921 < μ < 70.479.

User Pabloks
by
8.4k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories