Question

The manager of a department store believes that the mean annual income of the store's credit card customers is at least $21,000. For a sample of 90 customers the mean and standard deviation of income were found to be $20700 and $1450 respectively.

(1) Use the tables in the textbook to determine the critical value for the test statistic at the 5% level of significance. State your answer correct to three decimal places.

(2) Determine the calculated test statistic correct to 3 decimal places

Answer 1

Solution:

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

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H₀: The mean annual income of the store's credit card customers is at least $21,000.

Alternative hypothesis: H_a: The mean annual income of the store's credit card customers is less than $21,000.

H₀: µ ≥ 21000 versus H_a: µ < 21000

This is a lower tailed or left tailed test.

(1) Use the tables in the textbook to determine the critical value for the test statistic at the 5% level of significance. State your answer correct to three decimal places.

We are given level of significance = α = 0.05 or 5%

Sample size = n = 90

df = n – 1 = 90 – 1 = 89

Critical value = -1.662

(by using t-table)

(2) Determine the calculated test statistic correct to 3 decimal places

We are given

Xbar = 20700

S = 1450

n = 90

Test statistic formula is given as below:

t = (Xbar - µ) / [S/sqrt(n)]

t = (20700 – 21000) / [1450/sqrt(90)]

t = -300/ 152.8434

t = -1.9628

Test statistic = -1.963

P-value = 0.0264

(by using t-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is insufficient evidence to conclude that the mean annual income of the store's credit card customers is at least $21,000.