In: Statistics and Probability
Recovering from surgery: A new postsurgical treatment was compared with a standard treatment. Ten subjects received the new treatment, while ten others (the controls) received the standard treatment. The recovery times, in days, are given below.
Treatment: |
15 |
16 |
17 |
19 |
23 |
24 |
25 |
26 |
27 |
28 |
---|---|---|---|---|---|---|---|---|---|---|
Control: |
11 |
12 |
14 |
16 |
17 |
19 |
20 |
21 |
22 |
23 |
Send data to Excel |
Can you conclude that the mean recovery time for those receiving the new treatment is greater than the mean for those receiving the standard treatment? Let
μ1 denote the mean recovery time for the new treatment. Use the =α0.10 level of significance and the TI-84 Plus calculator.
Part 1 of 4
Your Answer is correct State the null and alternate hypotheses.
:H0 |
=μ1μ2 |
:H1 |
>μ1μ2 |
The hypothesis test is a ▼right-tailed test.
Part 2 of 4
Your Answer is incorrect
Compute the P-value. Round the answer to at least four decimal places.
P -value =0.0144 |
(this p-value was incorrect- it is not 0.0144) I need the correct answer and if it should reject/ do not reject hypothesis
Here, two different groups are used to collect data in two different situations. Further we do not know population standard deviation (or variance). So, we have to perform two sample t-test.
Suppose, random variables X and Y denote time required in new and standard treatments respectively.
Null hypothesis
Alternative hypothesis
Clearly, this ia a right tailed test.
The EXCEL outputs are as follows.
Degrees of freedom
This P-value can be calculated in TI-84 calculator using
tcdf(2.223899,999,18)
Actually 999 is used as a large number. Any sufficiently large number can be used in place of it.
Level of significance
We reject our null hypothesis if
Here, we observe that
So, we reject our null hypothesis.
Hence, based on the given data we can conclude that standard treatment takes significantly more time to recover than new treatment.