Question

Given that a sample of 100 values has a mean is 78.3 and the population

standard deviation is 5.7 create an 80% confidence interval for the true mean.

Answer 1

An 80% confidence interval for the true mean, using a sample mean of 78.3, population standard deviation of 5.7, and sample size of 100, results in the interval (77.56926, 79.03074).

Step-by-step explanation:

To create an 80% confidence interval for the true mean, we utilize the sample mean, the population standard deviation, and the Z-score for the desired confidence level.

With a given sample mean of 78.3, a population standard deviation of 5.7, and a sample size of 100, we look up the Z-score that corresponds to an 80% confidence level. The Z-score for an 80% confidence interval is approximately 1.282.

The formula for the confidence interval is:

Confidence Interval = Sample Mean ± (Z-score * (Population Standard Deviation / sqrt(Sample Size)))

Substituting the values we have:

Confidence Interval = 78.3 ± (1.282 * (5.7 / sqrt(100)))

Calculating further:

Margin of Error = 1.282 * (5.7 / 10) = 1.282 * 0.57 = 0.73074

Therefore, the confidence interval is:

78.3 ± 0.73074

Which gives us the interval:

(77.56926, 79.03074)

We can now say with 80% confidence that the true population mean lies within the interval (77.56926, 79.03074).