Question #151354
Question 5
1. 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.
1
Expert's answer
2020-12-18T11:22:00-0500

a. The following null and alternative hypotheses need to be tested:

H0:μ400H_0:\mu\leq400

H1:μ>400H_1:\mu>400

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.


b. Based on the information provided, the significance level is α=0.05,\alpha=0.05, and the number of  degrees of freedom is df=251=24.df=25-1=24. The critical value for a right-tailed test is tc=1.710882t_c=1.710882


c. The rejection region for this right-tailed test is R={t:t>1.710882}.R=\{t:t>1.710882\}.

The t-statistic is computed as follows:


t=Xˉμσ/n=41040020/25=2.5t=\dfrac{\bar{X}-\mu}{\sigma/\sqrt{n}}=\dfrac{410-400}{20/\sqrt{25}}=2.5

Since it is observed that t=2.5>1.710882=tc,t=2.5>1.710882=t_c, it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean μ\mu is greater than 400, at the 0.05 significance level.


d. Using the P-value approach:

df=24,t=2.5df=24, t=2.5

From the t-value table for one-tailed test

df=24,t=2.492,P(t>2.492)=0.01df=24, t=2.492,P(t>2.492)=0.01

df=24,t=2.797,P(t>2.797)=0.005df=24, t=2.797,P(t>2.797)=0.005

Then

df=24,t=2.5,P(t>2.5)=0.009827df=24, t=2.5,P(t>2.5)=0.009827

The p-value is p=0.009827,p=0.009827, and since p=0.009827<0.05=α,p=0.009827<0.05=\alpha, it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean μ\mu is greater than 400, at the 0.05 significance level.


e. There is enough evidence to claim that the population mean μ\mu is greater than GHc400, at the 0.05 significance level.



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