describe how you would find how find how much grass seed is needed to cover a triangular plot of land
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Answer:
The p-value for her test is 0.0075.
Step-by-step explanation:
She will then perform a hypothesis test to determine if the sample mean daily revenue provides significant evidence to conclude that the population mean daily revenue is less than $900 when the advertising dollars are reduced.
At the null hypothesis, we test that the mean is less than 900, that is:
At the alternate hypothesis, we test that the mean is less than 900, that is:
The test statistic is:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
In which X is the sample mean, is the value tested at the null hypothesis, s is the standard deviation of the sample and n is the size of the sample.
900 is tested at the null hypothesis:
This means that
Sample of 64 days. The first co-owner observes a sample mean daily revenue of $892.50 with a sample standard deviation of $24 when the advertising dollars have been reduced.
This means that
Value of the test:
What is the p-value for her test?
The pvalue of the test is the probability of a sample mean below 892.5, which is a left-tailed test for t = -2.5 with 64 - 1 = 63 degrees of freedom.
Using a calculator, the p-value for her test is 0.0075.