Length of Hospital Stays: The average length of “short” hospital stays for men is slightly longer thatn that for women: 5.2 days vs. 4.5 days. A random sample of recent hospital stays for both men and women revealed the following. At ? = 0.01, is there sufficient evidence to conclude that the average hospital stay for men is longer than the average hospital stay for women?
MEN WOMEN
Sample Size 32 30
Sample Mean 5.5 days 4.2 days
Pop. Standard Deviation 1.2 days 1.5 days
To determine if there is sufficient evidence to conclude that the average hospital stay for men is longer than the average hospital stay for women, we need to conduct a hypothesis test.
First, let's define the null and alternative hypotheses:
Null Hypothesis (H0): The average hospital stay for men is not longer than the average hospital stay for women.
Alternative Hypothesis (Ha): The average hospital stay for men is longer than the average hospital stay for women.
Since we want to compare the means of two independent samples, we can use the independent samples t-test.
Next, we can calculate the test statistic and the p-value:
1. Calculate the pooled standard deviation using the formula:
Sp = sqrt(((n1-1)*s1^2 + (n2-1)*s2^2) / (n1 + n2 - 2))
where n1 and n2 are the sample sizes, and s1 and s2 are the sample standard deviations.
Let's plug in the values:
n1 = 32, s1 = 1.2
n2 = 30, s2 = 1.5
Sp = sqrt(((32-1)*1.2^2 + (30-1)*1.5^2) / (32 + 30 - 2))
= sqrt((31*1.44 + 29*2.25) / 60)
= sqrt((44.64 + 65.25) / 60)
= sqrt(109.89 / 60)
= sqrt(1.8315) approx. 1.35 days
2. Calculate the test statistic (t-value) using the formula:
t = (x1 - x2) / (Sp * sqrt(1/n1 + 1/n2))
where x1 and x2 are the sample means, Sp is the pooled standard deviation, and n1 and n2 are the sample sizes.
Let's plug in the values:
x1 = 5.5, x2 = 4.2, Sp = 1.35, n1 = 32, n2 = 30
t = (5.5 - 4.2) / (1.35 * sqrt(1/32 + 1/30))
= 1.3 / (1.35 * sqrt(0.03125 + 0.03333))
= 1.3 / (1.35 * sqrt(0.06458))
= 1.3 / (1.35 * 0.2541)
= 1.3 / 0.3434
= 3.79 (rounded to two decimal places)
3. Find the p-value associated with the test statistic using the t-distribution table or a statistical software.
At a significance level of α = 0.01, we are looking for the p-value.
4. Compare the obtained p-value to the significance level. If the p-value is less than α, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
So, at α = 0.01, compare the p-value to 0.01. If the p-value is less than 0.01, we have sufficient evidence to conclude that the average hospital stay for men is longer than the average hospital stay for women. If the p-value is greater than or equal to 0.01, we do not have sufficient evidence.
Please note that step 3 requires the use of a t-distribution table or statistical software to find the p-value associated with the calculated t-value.
To test for the difference in the average hospital stays for men and women, we can use a two-sample t-test.
Step 1: State the null and alternative hypotheses:
Null hypothesis (H0): The average hospital stay for men is not longer than the average hospital stay for women.
Alternative hypothesis (Ha): The average hospital stay for men is longer than the average hospital stay for women.
Step 2: Determine the significance level (α):
The significance level (α) is given as 0.01.
Step 3: Compute the test statistic:
The formula for the test statistic for two-sample t-test comparing means is:
t = (x1 - x2) / sqrt((s1^2 / n1) + (s2^2 / n2))
Where:
x1 = sample mean for men
x2 = sample mean for women
s1 = population standard deviation for men
s2 = population standard deviation for women
n1 = sample size for men
n2 = sample size for women
Plugging in the given values:
x1 = 5.5 days
x2 = 4.2 days
s1 = 1.2 days
s2 = 1.5 days
n1 = 32
n2 = 30
t = (5.5 - 4.2) / sqrt((1.2^2 / 32) + (1.5^2 / 30))
Step 4: Calculate the degrees of freedom:
The degrees of freedom (df) for a two-sample t-test is given by:
df = (s1^2 / n1 + s2^2 / n2)^2 / [( (s1^2 / n1)^2 / (n1 - 1) ) + ( (s2^2 / n2)^2 / (n2 - 1) )]
Plugging in the given values:
df = ( (1.2^2 / 32) + (1.5^2 / 30) )^2 / ( ( (1.2^2 / 32)^2 / (32 - 1) ) + ( (1.5^2 / 30)^2 / (30 - 1) ) )
Step 5: Find the critical value:
Since we are testing if the average hospital stay for men is longer, we'll do a one-tailed test.
Using a t-table or statistical software, look up the critical value for a one-tailed test with a significance level (α) of 0.01 and the calculated degrees of freedom.
Step 6: Compare the test statistic with the critical value:
If the test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Step 7: Draw a conclusion:
Based on the results from step 6, provide a conclusion in the context of the problem.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.