Answer:
Explanation:
We would use the t- distribution.
From the information given,
Mean, μ = 34 inches
Standard deviation, σ = 7 inches
number of sample, n = 15
Degree of freedom, (df) = 15 - 1 = 14
Alpha level,α = (1 - confidence level)/2
α = (1 - 0.99)/2 = 0.005
We will look at the t distribution table for values corresponding to (df) = 14 and α = 0.005
The corresponding z score is 2.977
We will apply the formula
Confidence interval
= mean ± z ×standard deviation/√n
It becomes
34 ± 2.977 × 7/√15
= 34 ± 2.977 × 1.81
= 34 ± 5.39
The lower end of the confidence interval is 34 - 5.39 =28.61
The upper end of the confidence interval is 34 + 5.39 =39.39