Respuesta :
Answer:
The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.
Step-by-step explanation:
20-year mean snowfall in the Denver/Boulder region is 28.76 inches. Test if the snowfall for the 1993-1994 winters has higher than the previous 20-year average.
At the null hypothesis, we test if the average was the same, that is, of 28.76 inches. So
[tex]H_0: \mu = 28.76[/tex]
At the alternate hypothesis, we test if the average incresaed, that is, it was higher than 28.76 inches. So
[tex]H_1: \mu > 28.76[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
28.76 is tested at the null hypothesis:
This means that [tex]\mu = 28.76[/tex]
Standard deviation of 7.5 inches. However, for the winter of 1993-1994, the average snowfall for a sample of 32 different locations was 33 inches.
This means that [tex]\sigma = 7.5, X = 33, n = 32[/tex].
Value of the test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{33 - 28.76}{\frac{7.5}{\sqrt{32}}}[/tex]
[tex]z = 3.2[/tex]
P-value of the test and decision:
The p-value of the test is the probability of finding a sample mean above 33, which is 1 subtracted by the p-value of z = 3.2. In this question, we consider the standard level [tex]\alpha = 0.05[/tex].
Looking at the z-table, z = 3.2 has a p-value of 0.9993.
1 - 0.9993 = 0.0007
The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.
