Question

Men are believed to have a slightly longer average length of short hospital stays than women; 5.2 days versus 4.5 days respectively. A random sample of short hospital stays for men and women showed the following. At ???? =0.01, is there enough evidence to conclude the average short hospital stay for men is indeed longer than that for women? (Use the traditional method to test the hypothesis).

Answer 1

Answer:

Since our calculated value is lower than our critical value,$z_{calc}=2.02$, we have enough evidence to FAIL to reject the null hypothesis at 1% of singificance. So there is not nough evidence to conclude that mean for men it's significantly higher than the mean for female.

Step-by-step explanation:

1) Data given and notation  

$\bar X_{1}=5.2$ represent the mean for group men  

$\bar X_{2}=4.5$ represent the mean for group women  

Assuming these values for the remaining data:

$\sigma_{1}=1.2$ represent the population standard deviation for the sample men

$\sigma_{2}=1.5$ represent the population standard deviation for the sample women

$n_{1}=32$ sample size for the group men  

$n_{2}=30$ sample size for the group women  

z would represent the statistic (variable of interest)

$p_v$ represent the p value  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for men it's higher than the mean for women, the system of hypothesis would be:  

H0:$\mu_{1} \leq \mu_{2}$  

H1:$\mu_{1} > \mu_{2}$  

If we analyze the size for the samples both are higher than 30, and we know the population deviation's, so for this case is better apply a z test to compare means, and the statistic is given by:  

$z=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)  

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

3) Calculate the statistic  

We have all in order to replace in formula (1) like this:  

$z=\frac{5.2-4.5}{\sqrt{\frac{1.2^2}{32}+\frac{1.5^2}{30}}}=2.02$  

4) Find the critical value

In order to find the critical value we need to take in count that we are conducting a right tailed test, so we are looking on the normal standard distribution a value that accumulats 0.01 of the area on the right and 0.99 of the area on the left. We can us excel or a table to find it, for example the code in Excel is:

"=NORM.INV(1-0.01,0,1)", and we got $z_{critical}=2.33$

5) Statistical decision

Since our calculated value is lower than our critical value,$z_{calc}=2.02$, we have enough evidence to FAIL to reject the null hypothesis at 1% of significance. So there is not nough evidence to conclude that mean for men it's significantly higher than the mean for female.