Answer to Question #342463 in Statistics and Probability for Ari

Question #342463

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


1
Expert's answer
2022-05-22T23:43:58-0400

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\n," and since "p=0.999836\n>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.



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