An IQ is designed so that the mean is 100 and the standard deviation is 23 for the population of

Normal adults, find the sample size necessary to estimate the mean IQ score of statistics students
Such that it can be said with 90% confidence that the sample mean is within 5 IQ points of the true
Mean. Assume that @=23 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation.
The required sample size is --------- (round up to the nearest integer)

Why did the statistics student go broke?

Because they spent all their money on standard deviations!

To find the required sample size, we can use the formula:

n = (Z * σ / E)^2

Where:
n is the required sample size
Z is the Z-score for the desired confidence level (90% confidence corresponds to Z = 1.645)
σ is the population standard deviation (23 IQ points)
E is the desired margin of error (5 IQ points)

Let's plug in the values:

n = (1.645 * 23 / 5)^2

n = (37.935 / 5)^2

n = 7.587^2

n ≈ 57.60

Rounding up to the nearest integer, the required sample size is 58.

Now, is this a reasonable sample size for a real-world calculation? Well, it depends on various factors such as resources, time constraints, and the size of the population. 58 is a relatively small sample size, so it might not capture the complete variability of the statistics student population. However, if the population is relatively homogeneous and resources are limited, a sample size of 58 could still yield reasonably reliable results.

To determine the required sample size needed for estimating the mean IQ score of statistics students, we can use the formula for sample size calculation with a given margin of error and confidence level.

The formula for sample size calculation in estimating a population mean is:
n = (Z * σ / E)^2

Where:
n = required sample size
Z = Z-value for the desired confidence level
σ = population standard deviation
E = desired margin of error

In this case, we want to estimate the mean IQ score of statistics students with a 90% confidence level and a margin of error of 5 IQ points. The population standard deviation is given as 23.

Step 1: Find the Z-value for a 90% confidence level
To calculate the Z-value, we can use statistical tables or use technology. Let's assume we use technology, such as a statistical software or an online calculator, to find the Z-value for a 90% confidence level.

Using technology, we find that the Z-value for a 90% confidence level is approximately 1.645.

Step 2: Plug the values into the formula and calculate the required sample size (n)
n = (Z * σ / E)^2
n = (1.645 * 23 / 5)^2
n = (37.835 / 5)^2
n = 7.567^2
n ≈ 57.263

Step 3: Round up the result to the nearest integer
Since we cannot have a fraction of a sample, we round up the result to the nearest whole number. Thus, the required sample size is 58.

Therefore, to estimate the mean IQ score of statistics students with a 90% confidence level and a margin of error of 5 IQ points, a sample size of 58 is required.

Now, let's determine if this sample size is reasonable for a real-world calculation. The required sample size of 58 is determined based on the desired confidence level, margin of error, and population standard deviation. In general, a larger sample size tends to increase the precision and accuracy of the estimate. However, the overall reasonableness of the sample size also depends on other factors such as the available resources, budget, time, and practicality.

It is advisable to consult with a statistician or use statistical conventions specific to the field of study to assess if a sample size of 58 is appropriate for the study of statistics students' IQ scores.