The management of Regional Hospital has made substantial improvements in their h
ID: 3258507 • Letter: T
Question
The management of Regional Hospital has made substantial improvements in their hospital and would like to test and determine whether there has been a significant decrease in the average length of stay of their patients in their hospital. The following data has been accumulated from before and after the improvements. At 95% confidence, test to determine if there has been a significant reduction in the average length of stay.
After
Before
Sample size
45
56
Mean (in days)
4.6
4.9
Standard deviation
0.5
0.55
PLEASE SHOW ALL WORK!
a. Formulate the hypotheses
b. Compute the test statistic.
After
Before
Sample size
45
56
Mean (in days)
4.6
4.9
Standard deviation
0.5
0.55
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: Before < After
Alternative hypothesis: Before > After
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 0.1047
DF = 99
t = [ (x1 - x2) - d ] / SE
t = - 2.87
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.
The observed difference in sample means produced a t statistic of - 2.87. We use the t Distribution Calculator to find P(t < - 2.87) = 0.00251.
Therefore, the P-value in this analysis is 0.00251
Interpret results. Since the P-value (0.00251) is less than the significance level (0.05), we have to reject the null hypothesis.
From the above test we have sufficient evidence in the favor of the claim that if the average length of stay after improvements is significantly less than the average length of stay before.
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