Question

A study by a bank compared the mean savings of customers who were depositors for three years or less, with those who had been depositors for more than three years. The results of a sample are:

Depositors 3 years or less, sample size=100, mean saving balance=$1200, Population standard variation=$100

Depositors more than 3 years, sample size=150, mean saving balance=$1250, Population standard variation=$250

At the 0.10 significance level, can we conclude that those who had been depositing for more than three years carry higher balances than those who had been depositing three years or less? What is the p-value? Complete steps below.  

A. State the null hypothesis and the alternate hypothesis.

B. State the level of significance.

C. State the test statistic.

D. Formulate the decision rule.

E. Make a decision. Please show your work.

F. Determine the p-value (please show your work); Interpret the result of the hypothesis test.

Answer 1

Given values: (1) The provided sample means are : X̅1 = 1200.0 X̅2 = 1250.0 and the known population standard deviations are : σ1 = 100.0 σ2 = 250.0 and the sample size are n1 = 100.0 and n2 = 150.0. (2) Our test hypothesis is : The following null and alternative hypotheses need to be tested, hypothesis = h0: μ1 = μ2 vs H1: μ1 < μ1 μ1: mean of deposited less then 3 year μ2: mean of deposited more then 3 year This hypothesis corresponds to a lower tailed, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (3) Test statistics : The calculation of the z-test proceeds as follows, z = X̅1 - X̅2 - Δ 1200.0 - 1250.0 - 0 ---------------------- = -------------------------------- = -2.1997 √(σ1²/n1)+(σ2²/n2) √(100.0²/100.0) + (250.0²/150.0) (4) Rejection Criteria : Based on the information provided, the significance level is α = 0.1,and the degrees of freedom are df= 248.0 and the critical value for a lower tailed test is z tabulated = 1.2816 And the rejection region for this upper-tailed test is R =[z: z < -1.2816] (5) Decision about the null hypothesis : Since it is observed that z calculated means z = -2.1997 < z tabulated = -1.2816 it is then concluded that the null hypothesis is Rejected. Using the P-value approach: The p-value 0.01391, and since p = 0.01391 > α = 0.1 it is then concluded that the null hypothesis is Rejected. (6) Conclusion It is concluded that the null hypothesis Ho is Rejected.. Therefore, there is enough evidence to claim that population μ2 is greater than μ1 , at the 0.1 significance level.