Scores on the ACT college entrance exam are Normally distributed with a mean µ = 20 and standard deviation σ = 5. An SRS of 25 ACT scores is selected, construct a 95% Confidence Interval for the sample mean (X-Bar) and interpret your result.

95% = mean ± 1.96 SEm

SEm = SD/√n

I'll let you do the calculations and interpretations.

To construct a confidence interval for the sample mean, we can use the formula:

Confidence Interval = X-Bar ± (Z * (σ / √(n)))

Where:
X-Bar = sample mean
Z = z-score corresponding to the desired confidence level
σ = population standard deviation
n = sample size

In this case, the sample mean (X-Bar) and the population standard deviation (σ) are given. We are also given that we want to construct a 95% confidence interval.

To find the z-score corresponding to a 95% confidence level, we can use a standard normal distribution table or a calculator. For a 95% confidence level, the z-score is approximately 1.96.

Next, we substitute the given values into the formula:

Confidence Interval = 20 ± (1.96 * (5 / √(25)))

Simplifying the formula:

Confidence Interval = 20 ± (1.96 * (5 / 5))

Confidence Interval = 20 ± (1.96 * 1)

Confidence Interval = 20 ± 1.96

Finally, we can interpret the result. The confidence interval for the sample mean is (18.04, 21.96). This means that we are 95% confident that the true population mean falls within this interval. In other words, if we were to take many samples of size 25 and calculate confidence intervals, approximately 95% of these intervals would contain the true population mean.