Over the last 15 Major League Baseball seasons, the mean number of strikeouts by the American League leader is 258.5. Assuming that the number of strikeouts by the league leader is normally distributed and the standard deviation for all seasons in all leagues is 34.9, find the 82% Confidence Interval and the 93% Confidence Interval for the mean.

Question

Over the last 15 Major League Baseball seasons, the mean number of strikeouts by the American League leader is 258.5. Assuming that the number of strikeouts by the league leader is normally distributed and the standard deviation for all seasons in all leagues is 34.9, find the 82% Confidence Interval and the 93% Confidence Interval for the mean.

I got as far as 1-.82=.18 & 1-.93=.07. However, I'm not sure I have even that part right. Plz help!

Answer 1

To find the confidence intervals, you need to use the standard normal distribution (also known as Z-distribution) and apply a formula to calculate the intervals.

Let's start by finding the Z-scores for the given confidence levels: 82% and 93%.

For the 82% confidence level, the area under the standard normal curve will be (1 - 0.82) / 2 = 0.09 on each tail. Using a Z-table or a Z-score calculator, you can find the Z-score associated with a cumulative probability of 0.09. The Z-score turns out to be approximately 1.405.

For the 93% confidence level, the area under the standard normal curve is (1 - 0.93) / 2 = 0.035. With the same approach, the Z-score corresponding to a cumulative probability of 0.035 is approximately 1.812.

Now you can calculate the confidence intervals using the Z-scores, the sample mean, the standard deviation, and the sample size.

For the 82% Confidence Interval:
Confidence Interval = sample mean ± (Z-score * (standard deviation / sqrt(sample size)))

Confidence Interval = 258.5 ± (1.405 * (34.9 / sqrt(sample size)))

For the 93% Confidence Interval:
Confidence Interval = sample mean ± (Z-score * (standard deviation / sqrt(sample size)))

Confidence Interval = 258.5 ± (1.812 * (34.9 / sqrt(sample size)))

However, in your question, you have not provided the sample size. The formula for the confidence intervals requires the sample size to be specific, as it affects the precision of the intervals. Once you provide the sample size, you can substitute it into the formulas to calculate the confidence intervals.

For example, if the sample size were 100, the calculations would be as follows:

For the 82% Confidence Interval:
Confidence Interval = 258.5 ± (1.405 * (34.9 / sqrt(100)))

For the 93% Confidence Interval:
Confidence Interval = 258.5 ± (1.812 * (34.9 / sqrt(100)))

In both cases, you will obtain the lower and upper bounds of the confidence intervals.