Answer:
Step-by-step explanation:
We would use the t- distribution.
From the information given,
Mean, μ = 2950
Standard deviation, σ = 115
number of sample, n = 25
Degree of freedom, (df) = 25 - 1 = 24
Alpha level,α = (1 - confidence level)/2
α = (1 - 0.98)/2 = 0.01
We will look at the t distribution table for values corresponding to (df) = 24 and α = 0.01
The corresponding z score is 2.492
We will apply the formula
Confidence interval
= mean ± z ×standard deviation/√n
It becomes
2950 ± 2.492 × 115/√25
= 2950 ± 2.492 × 23
= 2950 ± 57.316
The lower end of the confidence interval is 2950 - 57.316 =2892.68
The upper end of the confidence interval is 2950 + 57.316 = 3007.32
The solution is correct.
