Question

The controller of a large retail chain is concerned about a possible slowdown in payments by customers. The controller randomly selects a sample of 10 accounts, with the following ages (in days):

40, 45, 50, 65, 70, 75, 78, 80, 82, 85

The sample standard deviation (s) is 15.53.

The controller wants to determine if the population mean number of days that the company must wait to get paid exceeds 65, which is the historical average. If it does, the company must take some action to ensure they are paid in a more timely manner

Calculate the test statistic (round to three decimal places, e.g. 2.0155 becomes 2.016).

Select the most appropriate critical value.

Answer 1

Given that, sample size (n) = 10

sample standard deviation (s) = 15.53

sample mean is,

The null and alternative hypotheses are,

H0 : μ = 65

Ha : μ > 65

Test statistic is,

=> Test statistic = t = 0.407

Degrees of freedom = 10 - 1 = 9

We used significance level = 0.05, since, we don't have the given significance level.

t-critical value at significance level of 0.05 with 9 degrees of freedom is, t* = 1.833

Excel Command : =TINV(0.10, 9) = 1.833

=> Critical value = 1.833

Note : if we don't use given sample standard deviation then by using TI-83 plus calculator we get, test statistic = 0.386