Answer to Question #342463 in Statistics and Probability for Ari

Question

The CEO of Global Investment Funds claims that, on average, clients leave their investment funds with the company for a period of at least 56 months. Test this claim at the 10% significance level, if it was found that a sample of 29 clients left their investment funds with the company for an average period of 58.85 months, with a standard deviation of 3.75 months

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge56$

$H_1:\mu<56$

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

Based on the information provided, the significance level is $\alpha = 0.1,$ $df=n-1=28$ degrees of freedom, and the critical value for a left-tailed test is $t_c = -1.312527.$

The rejection region for this left-tailed test is $R = \{t:t<-1.312527\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{58.85-56}{3.75/\sqrt{29}}\approx4.0927$$

Since it is observed that $t=4.0927>-1.312527=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for left-tailed, $df=28$ degrees of freedom, $t=4.0927,$ is $p= 0.999836 ,$ and since $p=0.999836 >0.1=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is less than 56, at the $\alpha = 0.1$ significance level.