Answer to Question #246042 in Statistics and Probability for boo

Question

A journal article reported that the mean hospital stay following a particular surgical procedure
in 2001 was 7.1 days. A researcher feels that the mean hospital stay in 2002 should be less due
to initiatives aimed at reducing health care costs. A random sample of 40 patients undergoing
the same surgical procedure in 2002 had a mean length of stay of 6.85 days with a standard
deviation of 7.01 days. Run the appropriate statistical test at

Answer 1

Given that the mean hospital stay is \mu= 7.1 days and to test that in 2002 whether the stay will be less due to the initiative aimed at reducing the healthcare costs a sample of n = 40 patients is selected which results in sample mean stay of $\bar{X}= 6.85$ days sand the sample standard deviation is s = 7.01 days.

Thus based on the claim the hypotheses are:

$H_0: \mu = 7.1 \\ H_1 : \mu < 7.1$

Based on the hypothesis it will be a left tailed test.

Since the sample size is greater than 30 and the population standard deviation is unknown hence t-distribution is applicable for hypothesis testing. So, a degree of freedom is used which is calculated as df = n-1= 40-1 =39.

Test statistic:

$t= \frac{\bar{X}- \mu}{s / \sqrt{n}} \\ t = \frac{6.85 -7.1}{7.01 / \sqrt{n}} = -0.226$

Rejection region:

Reject the H if P-value is less than 0.05.

P-value:

The P-value is computed using the excel formula for t-distribution which is =T.DIST(-0.226, 39, TRUE), thus the P-value is computed as 0.4114.

Conclusion: Since the P-value is greater than 0.05 hence we fail to reject the null hypothesis and conclude that there is insufficient evidence to support the claim that the mean days has decreased.