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. T
he critical value for a 98% confidence interval around this sample mean is ________.
1 answer:
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
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