6,859^1/3 can be written as
1 answer:
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Answer:
If we compare the p value and a significance level assumed we see that
so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the time is not significantly less than 20 minutes at 5% of significance.
Step-by-step explanation:
Data given and notation
represent the sample mean
represent the standard deviation for the sample
sample size
represent the value that we want to test
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 to be tested
We need to conduct a hypothesis in order to determine if the mean for the new forms take less time to complete than the older forms, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
Compute the test statistic
We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, 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 if the mean is (higher, less or not equal) to an specified value".
We can replace in formula (1) the info given like this:
Now we need to find the degrees of freedom for the t distirbution given by:
What do you conclude?
Compute the p-value
Since is a one left tailed test the p value would be:
If we compare the p value and a significance level assumed we see that
so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the time is not significantly less than 20 minutes at 5% of significance.