Question

A sample of 64 account balances from a credit company showed an average daily balance of $1,040. The standard deviation of the population is known to be $200. We are interested in determining if the mean of all account balances (i.e., population mean) is significantly different from $1,000.

Answer 1

Answer:

With 95% of confidence, there is no statistical evidence to confirm that the mean of all account balances is different from $1,000.

Explanation:

Null hypothesis (H0): μ=1000  

Alternative hypothesis (H1): μ≠1000  

We must use the z distribution because we know the population standard deviation.

z-statistic formula:  

z= (xbar-m)/(σ/(sqrt(n)))  

xbar: sample mean  

m: hypothesized value  

σ: population standard deviation  

n: number of observations  

z=(1,040-1,000)/(200/sqrt(64))  

z-statistic= 1.6

Because the problem do not specify the significance level, we will use 5%

The critical value from the z distribution with, 64-1 degrees of freedom and 2.5% significance level (because is a two-tail test) , is 1,96.

Because the z-statistic is less than the critical value, there is no statistical evidence to confirm that the mean of all account balances is different from $1,000.