Answer:
Step-by-step explanation:
We want to determine a 95% confidence interval for the mean mean test score of students.
Number of sample, n = 25
Mean, u = 81.5
Standard deviation, s = 10.2
For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.
We will apply the formula
Confidence interval
= mean ± z score ×standard deviation/√n
It becomes
81.5 ± 1.96 × 10.2/√25
= 81.5 ± 1.96/× 2.04
= 81.5 ± 3.9984
The lower end of the confidence interval is 81.5 - 3.9984 =77.5
The upper end of the confidence interval is 81.5 + 3.9984 = 85.5
Therefore, with 95% confidence interval, the mean test score of students is between 77.5 and 85.5
