Weatherwise is a magazine published by the American Meteorological Society. One issue gives a rating system used to classify Nor'easter storms that frequently hit New England and can cause much damage near the ocean. A severe storm has an average peak wave height of μ = 16.4 feet for waves hitting the shore. Suppose that a Nor'easter is in progress at the severe storm class rating. Peak wave heights are usually measured from land (using binoculars) off fixed cement piers. Suppose that a reading of 39 waves showed an average wave height of x = 16.5 feet. Previous studies of severe storms indicate that σ = 3.5 feet. Does this information suggest that the storm is (perhaps temporarily) increasing above the severe rating? Use α = 0.01.

Respuesta :

Answer:

pvalue = 0.4286 > 0.01, so this information does not suggest that the storm is (perhaps temporarily) increasing above the severe rating

Step-by-step explanation:

The null hypothesis is:

[tex]H_{0} = 16.4[/tex]

The alternate hypotesis is:

[tex]H_{1} > 16.4[/tex]

Our test statistic is:

[tex]t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}[/tex]

In which X is the statistic, [tex]\mu[/tex] is the mean, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.

In this problem, we have that:

[tex]\mu = 16.4, \sigma = 3.5, X = 16.5, n = 39[/tex]

So

[tex]t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}[/tex]

[tex]t = \frac{16.5 - 16.4}{\frac{3.5}{\sqrt{39}}}[/tex]

[tex]t = 0.18[/tex]

Looking at the z-table, z = t = 0.18 has a pvalue of 0.5714.

The alternate hypothesis is accepted if there is a lower than 1%(Because of α = 0.01) probability of finding a value higher than X.

In this problem, X = 16.5, with a pvalue of 0.5714.

1 - 0.5714 = 0.4286 > 0.01

So this information does not suggest that the storm is (perhaps temporarily) increasing above the severe rating