Question

Bob's average golf score at his local course is 93.6 from a random sample of 34 rounds of golf. Assume that the population standard deviation for his golf score is 4.2. The 90% confidence interval around this sample mean is:_______.

a. (92.4, 94.8).
b. (89.0, 98.3).
c. (90.5, 96.7).
d. (93.2, 94.0).

Answer 1

a. (92.4, 94.8)

Explanation:

Given that:

Mean (m) = 93.6

Standard deviation (σ) = 4.2

Sample size (n) = 34

α = 90%

Confidence interval around sample mean :

Mean ± Zstatistic * s/√n

Z0.90 = 1.645 (Z probability calculator)

s / √n = 4.2 / √34 = 0.7202940

Lower limit = 93.6 - (1.645 * 0.7202940) = 92.41511637 = 92.4

Upper limit = 93.6 + (1.645 * 0.7202940) = 94.78488363 = 94.8