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

standard deviation of 4.5 on a placement test. Find a 95% confidence interval
for the mean score, ì, of all students taking the test. (Note: ó is unknown.)

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

CI = X̄ ± t * (s / √n)

Where:
- CI represents the confidence interval
- X̄ is the sample mean (30.8 in this case)
- t represents the critical value (we need to find this)
- s is the sample standard deviation (4.5 in this case)
- n is the sample size (56 in this case)
- ± indicates the range of the confidence interval

To find the critical value (t) for a 95% confidence level, we need to determine the degrees of freedom (df) and consult the t-distribution table. The degrees of freedom can be calculated using the formula df = n - 1.

In this case, df = 56 - 1 = 55.

Next, we need to find the corresponding t-value for a 95% confidence level and df = 55 in the t-distribution table. Let's assume a 2-tailed test since we want to find the confidence interval.

Consulting the t-distribution table, the t-value for a 95% confidence level and df = 55 is approximately 2.009.

Now we can substitute the values into the formula:

CI = 30.8 ± 2.009 * (4.5 / √56)

Calculating the expression inside the parentheses:

CI = 30.8 ± 2.009 * (4.5 / 7.483)

Simplifying the expression:

CI = 30.8 ± 2.009 * 0.6019

Calculating the multiplication:

CI = 30.8 ± 1.208

Calculating the lower and upper limits of the confidence interval:

Lower limit = 30.8 - 1.208 = 29.592
Upper limit = 30.8 + 1.208 = 32.008

Therefore, the 95% confidence interval for the mean score μ is approximately 29.592 to 32.008.

To find the 95% confidence interval for the mean score, we can use the formula:

Confidence Interval = Mean ± (Z * Standard Deviation / √sample size)

First, let's calculate the Z-value for a 95% confidence level. We can use the standard normal distribution table or a calculator to find this value. For a 95% confidence level, the Z-value is approximately 1.96.

Now, we can substitute the given values into the formula:

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

To calculate the square root of 56, we get √56 = approximately 7.4833.

Confidence Interval = 30.8 ± (1.96 * 4.5 / 7.4833)

Next, we can calculate the value inside the parentheses:

(1.96 * 4.5 / 7.4833) ≈ 1.1806

Finally, substitute this value into the formula:

Confidence Interval = 30.8 ± 1.1806

Thus, the 95% confidence interval for the mean score, ì, is approximately:

30.8 - 1.1806 ≤ ì ≤ 30.8 + 1.1806

Therefore, the 95% confidence interval for the mean score, ì, is approximately:

29.6194 ≤ ì ≤ 32.9806