Respuesta :
Answer:
1) There is enough evidence to support the claim that the average hospital stay for men is longer than the average hospital stay for women. (P-value = 0.00051).
2) zc=2.33
Step-by-step explanation:
1) This is a hypothesis test for the difference between populations means.
The claim is that the average hospital stay for men is longer than the average hospital stay for women.
Then, the null and alternative hypothesis are:
H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2> 0
The significance level is 0.01.
The sample 1, of size n1=37 has a mean of 5.6 and a standard deviation of 1.2.
The sample 2, of size n2=35 has a mean of 4.5 and a standard deviation of 1.6.
The difference between sample means is Md=1.1.
[tex]M_d=M_1-M_2=5.6-4.5=1.1[/tex]
The estimated standard error of the difference between means is computed using the formula:
[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1.2^2}{37}+\dfrac{1.6^2}{35}}\\\\\\s_{M_d}=\sqrt{0.039+0.073}=\sqrt{0.112}=0.335[/tex]
Then, as we know the population standard deviation of both, we can calculate the z-statistic as:
[tex]z=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{1.1-0}{0.335}=\dfrac{1.1}{0.335}=3.29[/tex]
This test is a right-tailed test, with z=3.29, so the P-value for this test is calculated as:
[tex]P-value=P(z>3.29)=0.00051[/tex]
As the P-value (0.00051) is smaller than the significance level (0.01), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that the average hospital stay for men is longer than the average hospital stay for women.
2. The critical values for this right tail test are calculated from the z-table and the significance level (α=0.01)
[tex]P(z>z_c)=0.01[/tex]
This happens for zc=2.33:
[tex]P(z>2.33)=0.01[/tex]
