Answer to Question #304583 in Statistics and Probability for Alisia

Question

Medical aid board wants to find if the mean time to settle medical claims of members differs between 2 major medical schemes. The sample mean time to settlement was 10.8 days. Population standard deviation of 3.2 days for 1 and the other 12.4 days with population deviation 2.3 days

Formulate the null and alternative hypothesis and test this hypothesis at the 1% significance level and draw a conclusion?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1=\mu_2$

$H_1:\mu_1\not=\mu_2$

This corresponds to a two-tailed test, and a z-test for two means, with known population standard deviations will be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ and the critical value for a two-tailed test is $z_c=2.5758.$

The rejection region for this two-tailed test is $R = \{z: |z| > 2.5758\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{X}_1-\bar{X}_2}{\sqrt{\sigma^2_1/n_1+\sigma^2_2/n_2}}$$

$$=\dfrac{10.8-12.4}{\sqrt{3.2^2/14+2.3^2/15}}\approx-1.5367$$

Since it is observed that $|z| = 1.5367 \le 2.5758=z_c ,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=2P(Z<-1.5367)=0.124369,$ and since $p = 0.124369 \ge 0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu_1$ is different than $\mu_2,$ at the $\alpha = 0.01$ significance level.