Question

Researchers compare glucose concentrations in control old and dementia patients. Available data include:

complete/show work and equations used

Patient               n          Mean     Median          s (SD)                 

Dementia           37        8.7               9               2.03                   

Control              42         11.36           11              2.33                   

Perform a complete hypothesis test.

Answer 1

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​

Ha: μ1​ ≠ μ2​

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

The critical values are FL​=0.522 and FU​=1.889, and since F=0.759, then the null hypothesis of equal variances is not rejected.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=77. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this two-tailed test is tc​=1.991, for α=0.05 and df=77.

The rejection region for this two-tailed test is R={t:∣t∣>1.991}.

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

(4) Decision about the null hypothesis:Since it is observed that ∣t∣=2.449>tc​=1.991, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0166, and since p=0.0166<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion: It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1​ is different than μ2​, at the 0.05 significance level.