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