Answer:
$521.58 < μ < $666.1
Step-by-step explanation:
Spring break can be a very expensive holiday. A sample of 80 students is surveyed, and the average amount spent by students on travel and beverages is $593.84 at 92% confidence level. The sample standard deviation is approximately $369.34. Is $521.58 ≤μ≤ $666.10 correct?
Given that:
number of samples (n) = 80 students, mean (μ) = $593.84, standard deviation (σ) = $369.34, confidence level (c) = 92% = 0.92.
α = 1 - c = 1 - 0.92 = 0.08
the z score of 0.46 (0.5 - 0.04) is the same as the z score of 0.04. This is gotten from the Normal Distribution Table.
Therefore,
The margin of error (e) is given as:
The confidence interval = (μ - e, μ + e) = ($593.84 - $72.26, $593.84 + $72.26) = ($521.58, $666.1)
The confidence interval is $521.58 < μ < $666.1