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Complete question:
The manager of a supermarket would like to determine the amount of time that customers wait in a check-out line. He randomly selects 45 customers and records the amount of time from the moment they stand in the back of a line until the moment the cashier scans their first item. He calculates the mean and standard deviation of this sample to be barx = 4.2 minutes and s = 2.0 minutes. If appropriate, find a 90% confidence interval for the true mean time (in minutes) that customers at this supermarket wait in a check-out line
Answer:
(3.699, 4.701)
Step-by-step explanation:
Given:
Sample size, n = 45
Sample mean, x' = 4.2
Standard deviation = 2.0
Required:
Find a 90% CI for true mean time
First find standard error using the formula:
Standard error = 0.298
Degrees of freedom, df = n - 1 = 45 - 1 = 44
To find t at 90% CI,df = 44:
Level of Significance α= 100% - 90% = 10% = 0.10
Find margin of error using the formula:
M.E = S.E * t
M.E = 0.298 * 1.6802
M.E = 0.500938 ≈ 0.5009
Margin of error = 0.5009
Thus, 90% CI = sample mean ± Margin of error
Lower limit = 4.2 - 0.5009 = 3.699
Upper limit = 4.2 + 0.5009 = 4.7009 ≈ 4.701
Confidence Interval = (3.699, 4.701)