Asked by Vanessa Rodriguez on May 07, 2024
Verified
The manager of a grocery store has taken a random sample of 144 customers.The average length of time it took these 144 customers to check out was 3 minutes.It is known that the standard deviation of the population of checkout times is 1 minute.The 95% confidence interval for the true average checkout time (in minutes) is
A) 2.863 to 3.137.
B) 1.36 to 4.64.
C) 1 to 5.
D) 2.837 to 3.163.
Standard deviation of population
A measure that quantifies the variation or spread of a set of data points in a population.
- Develop an understanding of how to compute confidence intervals for means and proportions of a population.
Verified Answer
GS
Gurdeep SinghMay 08, 2024
Final Answer :
D
Explanation :
We can use the formula for a confidence interval of a population mean, which is:
CI = X̄ ± Zα/2 * σ/√n
Where:
- X̄ is the sample mean
- Zα/2 is the critical value for the desired confidence level (in this case, 95%). Zα/2 can be found using a standard normal distribution table or calculator. It is 1.96.
- σ is the population standard deviation
- n is the sample size
Substituting the given values, we have:
CI = 3 ± 1.96 * 1/√144
CI = 3 ± 0.16
CI = (2.84, 3.16)
Therefore, the 95% confidence interval for the true average checkout time is 2.837 to 3.163, which is closest to choice D.
CI = X̄ ± Zα/2 * σ/√n
Where:
- X̄ is the sample mean
- Zα/2 is the critical value for the desired confidence level (in this case, 95%). Zα/2 can be found using a standard normal distribution table or calculator. It is 1.96.
- σ is the population standard deviation
- n is the sample size
Substituting the given values, we have:
CI = 3 ± 1.96 * 1/√144
CI = 3 ± 0.16
CI = (2.84, 3.16)
Therefore, the 95% confidence interval for the true average checkout time is 2.837 to 3.163, which is closest to choice D.
Learning Objectives
- Develop an understanding of how to compute confidence intervals for means and proportions of a population.