Question

A sample of 64 account balances from a credit company showed an average daily balance of $1040. 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 $1000.

Which of the following is true

The Null Hypothesis can be rejected at the 2.5% significance level
	The Null Hypothesis can be rejected at the 10% significance level
	The Null Hypothesis cannot be rejected at the 5% significance level
	The Null Hypothesis cannot be rejected at the 20% significance level

Answer 1

Z test statistic formula

$Z=\frac{\bar{x}-\mu }{\frac{\sigma }{\sqrt{n}}}=\frac{1040-1000 }{\frac{200}{\sqrt{64}}}=1.6$

P value is 0.1096 ................................by using =2*(1-NORMSDIST(1.6)) in Excel.

P value is 0.1096 > 0.05

The Null Hypothesis cannot be rejected at the 5% significance level

Option C) is correct.