Question

A random sample of 17 hotels in Boston had an average nightly room rate of $165.40 with a sample standard deviation of $21.70. The critical value for a 98% confidence interval around this sample mean is ________.

Answer 1

Answer:

153.158 ; 177.642

Step-by-step explanation:

Given the following:

Sample mean (m) = 165.40

Sample standard deviation (s) = 21.70

Sample size (n) = 17

α = 98%

Confidence interval = m ± z(SE)

z at 98% = 2.326

SE = s/√n

SE = 21.70/√17 = 5.2630230

Hence,

Confidence interval = 165.40 ± 2.326(5.2630230)

165.40 - 12.241791498 OR 165.40 + 12.241791498

153.158 ; 177.642