Asked by Rhyana Delgado on May 02, 2024
Verified
A sample of 100 information systems managers had an average hourly income of $40.00 with a standard deviation of $8.00.The 95% confidence interval for the average hourly wage (in $) of all information system managers is
A) 38.684 to 41.316.
B) 32 to 48.
C) 38.432 to 41.568.
D) 39 to 41.
Confidence interval
A set of values obtained from sample data which probably encompasses the value of an unspecified population parameter.
Average hourly income
The mean income earned by an individual per hour of work.
Standard deviation
A metric that calculates the degree of spread or variability within a data set.
- Master the calculation and understanding of confidence intervals for the means and proportions of populations.
Verified Answer
GP
gianicola perronMay 06, 2024
Final Answer :
C
Explanation :
Using the formula for 95% confidence interval:
CI = X̄ ± Zα/2 * (σ/√n)
where:
X̄ = sample mean = $40
Zα/2 = 1.96 (for 95% confidence interval)
σ = population standard deviation = $8
n = sample size = 100
Substituting the given values:
CI = 40 ± 1.96 * (8/√100)
CI = 40 ± 1.568
CI = 38.432 to 41.568
Therefore, the 95% confidence interval for the average hourly wage of all information system managers is $38.432 to $41.568. Therefore, the correct choice is C.
CI = X̄ ± Zα/2 * (σ/√n)
where:
X̄ = sample mean = $40
Zα/2 = 1.96 (for 95% confidence interval)
σ = population standard deviation = $8
n = sample size = 100
Substituting the given values:
CI = 40 ± 1.96 * (8/√100)
CI = 40 ± 1.568
CI = 38.432 to 41.568
Therefore, the 95% confidence interval for the average hourly wage of all information system managers is $38.432 to $41.568. Therefore, the correct choice is C.
Learning Objectives
- Master the calculation and understanding of confidence intervals for the means and proportions of populations.