Question

A sample of 64 account balances from a credit company showed an average daily balance of $1145.

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 $1100. (In other words, we are testing the Null Hypothesis that the mean account balance for the population is $1100)

Which of the following is true

	The Null Hypothesis cannot be rejected at the 5% significance level
	The Null Hypothesis can be rejected at the 2.5% significance level
	The Null Hypothesis cannot be rejected at the 10% significance level
	The Null Hypothesis can be rejected at the 1% significance level
	The Null Hypothesis cannot be rejected at the 20% significance level

Answer 1

H0: = 1100

Ha: 1100

Test statistics

z = - / / sqrt(n)

= 1145 - 1100 / 200 / sqrt(64)

= 1.8

This is test statistics value.

p-value = 2 * P( Z > z)

= 2 * P( Z > 1.8)  

= 2 * 0.0359

= 0.0718

The null hypothesis is rejected if p-value < significance level .

Answer: The null hypothesis cannot be rejected at 5% level.