A group of 56 randomly selected students have a mean score of 30.8 and a

standard deviation of ó = 4.5 on a placement test. Find a 95% confidence interval
for the mean score, ì, of all students taking the test.

29.8 < m < 31.8

To find the 95% confidence interval for the mean score of all students taking the test, we can use the formula:

Confidence Interval = Mean ± (Critical Value) * (Standard Deviation / √n)

Where:
- Mean is the sample mean score (30.8 in this case)
- Critical Value is the value corresponding to the desired confidence level (95%), which can be found using a z-table or calculator
- Standard Deviation is the population standard deviation (4.5 in this case)
- n is the sample size (56 in this case)

Step 1: Finding the Critical Value
For a 95% confidence level, the critical value can be obtained by subtracting (1 - 0.95) / 2 from 1, which gives us a value corresponding to the upper 2.5% of the distribution. From the z-table, the critical value is approximately 1.96.

Step 2: Calculating the Confidence Interval
Plugging in the values into the formula, we have:

Confidence Interval = 30.8 ± (1.96) * (4.5 / √56)

Calculating the right side of the equation:

Confidence Interval = 30.8 ± (1.96) * (4.5 / 7.483)

Simplifying further:

Confidence Interval = 30.8 ± (1.96) * (0.602)

Finally, calculating the upper and lower limits of the confidence interval:

Confidence Interval ≈ 30.8 ± (1.18)

So, the 95% confidence interval for the mean score of all students taking the test is approximately (29.62, 32.98).