Suppose a random sample of 197 accounts from a corporate cre

Suppose a random sample of 197 accounts from a corporate credit card database revealsed a sample average balance of $2,325 with a standard deviation of $144. Use the sample information to develop a 95% confidence interval for the true population of all credit card balances for this corporate credit card.

Solution

sigma for a sampling distribution is sigma/sqrt(n)=144/sqrt(197)=10.25

95% confidence interval lies between mu-1.96sigma upto mu+1.96sigma

Hence it is

2325-(1.96*10.25)=2304.91

upto

2325+(1.96*10.25)=2345.09