The Deborah Heart Institute performs many open-heart surgery procedures. Recently research physicians at the Institute developed a new heart bypass procedure that they believe reduces the average length of recovery. The hospital board will not adopt the procedure unless there is substantial evidence to suggest that it is better than the current procedure. Records indicate that the mean recovery rate for the current procedure is 42 days with a standard deviation of 5 days. Test this hypothesis on a sample=36 patients and a sample mean of 40.2 days using a rejection region.
(can you provide complete answer with detailed steps)
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 related to the Z score.
To test this hypothesis, we will use a one-sample t-test. The null hypothesis (H0) is that the new heart bypass procedure does not reduce the average length of recovery, while the alternative hypothesis (Ha) is that the new procedure does reduce the average length of recovery.
Here are the steps to test the hypothesis:
Step 1: State the hypotheses:
- Null hypothesis (H0): The mean recovery rate for the new heart bypass procedure is equal to the mean recovery rate for the current procedure (µ = 42 days).
- Alternative hypothesis (Ha): The mean recovery rate for the new heart bypass procedure is less than the mean recovery rate for the current procedure (µ < 42 days).
Step 2: Set the significance level (α):
Let's assume a significance level of 0.05 (5%).
Step 3: Formulate the rejection region:
Since the alternative hypothesis is one-tailed (µ < 42 days), we need to find the critical t-value for a one-tailed test at a significance level of 0.05 and degrees of freedom (df) = sample size - 1.
- In this case, df = 36 - 1 = 35.
- Using a t-table or statistical software, we find the critical t-value to be approximately -1.6909.
Step 4: Calculate the test statistic:
- The test statistic for a one-sample t-test is given by:
t = (sample mean - null hypothesis mean) / (standard deviation / square root of sample size)
- Substituting the given values:
t = (40.2 - 42) / (5 / √36)
t = -1.8 / (5 / 6)
t = -1.8 * 6 / 5
t = -2.16
Step 5: Make a decision:
If the test statistic falls within the rejection region, we reject the null hypothesis; otherwise, we fail to reject it.
In this case, the calculated test statistic (-2.16) is smaller than the critical t-value (-1.6909). So, the test statistic falls within the rejection region.
Step 6: Interpret the result:
Since the test statistic falls within the rejection region, we reject the null hypothesis. This suggests that there is substantial evidence to suggest that the new heart bypass procedure reduces the average length of recovery compared to the current procedure for the patients at the Deborah Heart Institute.
In conclusion, based on the sample data, with a sample mean recovery rate of 40.2 days, we can reject the null hypothesis and conclude that the new heart bypass procedure has shown a statistically significant reduction in the average length of recovery compared to the current procedure.