Answer:
Since is a one sided test the p value would be:
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean time for the new treatment is significantly lower than the time for the standard treatment at 5% of significance.
Step-by-step explanation:
New: 12, 13,15,19,20,21
Standard: 18,23,24,30,32,35
We can calculate the sample mean and deviation with the following formulas:
Data given and notation
represent the mean for the new case
represent the mean for the standard case
represent the sample standard deviation for the new case
represent the sample standard deviation for the standard case
sample size selected for the standard case
sample size selected for the new case
represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if recovery time for patients who receive the new treatment is less than the mean for patients who receive the standard treatment, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:
(1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".
Calculate the statistic
We can replace in formula (1) the info given like this:
P-value
The first step is calculate the degrees of freedom, on this case:
Since is a one sided test the p value would be:
Conclusion
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean time for the new treatment is significantly lower than the time for the standard treatment at 5% of significance.
