Question

A bank executive believes that average bank balances for individuals equals $15,000. The executive takes a random sample of 25 individuals and finds a sample mean of $12,800 and a sample standard deviation (s) of $4,000. Test the hypothesis at the 0.05 and 0.01 levels.

Answer 1

Solution:

Here, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H0: The average bank balances for individuals equals $15,000.

Alternative hypothesis: Ha: The average bank balances for individuals is not equals $15,000.

H0: µ = 15000 versus Ha: µ ≠ 15000

This is a two tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 15000

Xbar = 12800

S = 4000

n = 25

df = n – 1 = 24

For α = 0.05

Critical value = - 2.0639 and 2.0639

(by using t-table or excel)

For α = 0.01

Critical value = - 2.7969 and 2.7969

(by using t-table or excel)

Test statistic is given as below:

t = (12800 – 15000)/[4000/sqrt(25)]

t = -2.75

P-value = 0.0111

(by using t-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is not sufficient evidence at α = 0.05 to conclude that the average bank balances for individuals equals $15,000.

P-value > α = 0.01

So, we do not reject the null hypothesis

There is sufficient evidence at α = 0.01 to conclude that the average bank balances for individuals equals $15,000.