In: Statistics and Probability
Please show work
5. Sears is in the process of hiring a new Manager for its Monmouth Mall store and they are trying to estimate the best annual healthcare premium. In a sample of 30 Managers, they found that the average yearly premium needed is $16,469 with a standard deviation of $2,000.
(a) What is the population mean? What is the best estimate of the population mean?
(b) Develop an 95% confidence interval for the population mean.
(c) Assuming a 99% level of confidence, how large of a sample is required with an acceptable error of $600?
(a) What is the population mean? What is the best estimate of the population mean?
The best estimate of the population mean is sample mean is sample mean $16469
(b) Develop an 95% confidence interval for the population mean.
The provided sample mean is 16469 and the sample standard deviation is s = 2000 The size of the sample is n = 30 and the required confidence level is 95%.
The number of degrees of freedom are df = 30 - 1 = 29 , and the significance level is α=0.05.
Based on the provided information, the critical t-value for α=0.05 and df = 29 degrees of freedom is t_c = 2.04
The 95% confidence for the population mean μ is computed using the following expression
Therefore, based on the information provided, the 95 % confidence for the population mean μ is
CI = (15722.188, 17215.812)
(c) Assuming a 99% level of confidence, how large of a sample is required with an acceptable error of $600?
z_c = 2.58 for 99% confidence interval
Acceptable error = E = 600
s = 2000
n = 73.96
n = 74