1. Suppose we’re interested in the average level of customer satisfaction. Show
ID: 2930404 • Letter: 1
Question
1. Suppose we’re interested in the average level of customer satisfaction. Show all your work and place a box around your final confidence interval.
a. Please construct a 95% confidence interval for the population average of customer satisfaction. Suppose we have questioned 25 customers and found the sample mean to be 6 and the sample standard deviation to be 2.
b. Please construct a 95% confidence interval for the population average of customer satisfaction. Suppose we have questioned 80 customers and found the sample mean to be 6 and the sample standard deviation to be 2.
c. Please construct a 90% confidence interval for the population average of customer satisfaction. Suppose we have questioned 30 customers and found the sample mean to be 6 and the sample standard deviation to be 2.
Explanation / Answer
(1-alpha)*100% confidence interval for population mean=sample mean±z(alpha/2)*sd/sqrt(n)
(a)95% confidence interval for population mean=sample mean±z(0.05/2)*sd/sqrt(n)
=6±1.96*2/sqrt(25)=6±0.784=(5.216, 6.784)
(b)95% confidence interval for population mean=sample mean±z(0.05/2)*sd/sqrt(n)
=6±1.96*2/sqrt(80)=6±0.44=(5.46, 6.44)
(c)90% confidence interval for population mean=sample mean±z(0.01/2)*sd/sqrt(n)
=6±1.645*2/sqrt(30)=6±0.6=(5.4, 6.6)
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