I've asked this before but I haven't gotten an answer how I need it :( I need to use the deffinition of the derivative to solve
this and it's getting complicated and I'm not exactly sure what to do. Anybody who knows how to solve it in that manner can you please help me
1 answer:
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Answer:
a.
The null hypothesis is .
The alternate hypothesis is .
b.
The test statistic is
c.
2. At α=0.05, there is sufficient evidence to conclude the mean delivery time exceeds 24 minutes.
Step-by-step explanation:
Question a:
Test if the mean delivery time actually exceeds 24 minutes.
At the null hypothesis, we test that the mean is of 24 minutes or less, that is:
At the alternate hypothesis, we test if it is more, that is:
Question b:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The test statistic is:
In which X is the sample mean, is the value tested at the null hypothesis,
is the standard deviation of the sample and n is the size of the sample.
24 is tested at the null hypothesis:
This means that
The owner has randomly selected 32 customers and has delivered food to their homes. The mean delivery time for the sample is 26 minutes with standard deviation of 5.8 minutes.
This means that .
Value of the test-statistic:
The test statistic is .
c. At α=0.05, which of the following conclusion is correct?
The p-value of the test is the probability of finding a sample mean above 26, which is a right-tailed test of t = 1.95 with 32 - 1 = 31 degrees of freedom.
Using a t-distribution calculator, this p-value is of 0.0301.
Since the p-value is 0.0301 < 0.05, there is sufficient evidence to conclude the mean delivery time exceeds 24 minutes, and the correct answer is given by option 2.