This is a difference of two means hypothesis test with the null and hypothesis tests as follows:
[tex]H_0:\mu_1=\mu_2 \\H_A:\mu_1\neq\mu_2[/tex]
The test statistics is given by
[tex]z= \frac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{ \sqrt{ \frac{\sigma_1^2}{n_1}+ \frac{\sigma_2^2}{n_2} } } [/tex]
where: [tex]\bar{x}_1=18.3[/tex]
[tex]\bar{x}_2=19.2[/tex]
[tex]\mu_1-\mu_2=0[/tex]
[tex]\sigma_1=2.1[/tex]
[tex]n_1=40[/tex]
[tex]\sigma_2=2.92[/tex]
[tex]n_2=50[/tex]
[tex]z= \frac{(18.3-19.2)-0}{ \sqrt{ \frac{(2.1)^2}{40}+ \frac{(2.92)^2}{50} } } \\ \\ = \frac{-0.9}{ \sqrt{0.11025+0.170528} } \\ \\ = \frac{-0.9}{ \sqrt{0.280778} } = \frac{-0.9}{0.5299} \\ \\ =-1.6985[/tex]
Since, the significance level is 0.02, the null hypothesis will be rejected if z < -2.33 or z > 2.33.
But, since z = -1.6985 which is not less than -2.33, we fail to reject the null hypothesis and we therefore, conclude that there is not enough evidence to conclude that there is a significant difference in the mean waiting time of patients for both hospitals.
