Question

Ecobank Ghana issues its own credit card. The credit manager, Giovina wants to find whether the mean monthly unpaid balance is not at most Ghc400. The level of significance is set at 0.05. A random check of 25 unpaid balances revealed that the average is GHc410 with the standard deviation of GHc20. Should the credit manager conclude that the population mean is not at most GHc400, or is it reasonable that the difference is due to chance? a. Explain why t is the test statistic to be used here. b. What is the critical value of the test statistic?

c. What is your decision regarding the H0? d. Estimate the p-value and use it to decide. e. Should the credit manager conclude that the population mean is not at most GHc400, or is it reasonable that the difference is due to chance? Explain briefly

Answer 1

Here we have:

$\bar x =410$

$s=20$

$n=25$

significance level $\alpha=0.05$

$H_0:\mu\le400$

$H_a:\mu>400$

The test is right-tailed.

a) Since the population standard deviation is unknown and the sample size is smaller than 30 we use t statistic.

b) $df=n-1=24$

The critical value for $\alpha = 0.05$ and 24 degrees of freedom is 1.711.

The critical region is $t>1.711$

c) Test staticstic:

$$t =\frac{\bar x-\mu}{\frac{s}{\sqrt n}}=\frac{410-400}{\frac{20}{\sqrt {25}}}=2.5$$

Since 2.5 > 1.711 thus t falls in the rejection region we reject the null hypothesis.

d) Then looking up in the Student's t-distribution table with 24 degrees of freedom we find the corresponding probability or p-value.

p-value is 0.0098

Since 0.0098 < 0.05 thus p-value is less than the significance level we reject the null hypothesis.

e) At the 5% significance level the data do provide sufficient evidence to support the claim. The credit manager is 95% confident to conclude that the population mean is not at most GHc400.