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 ? = 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 34 waves showed an average wave height of x = 17.3 feet. Previous studies of severe storms indicate that ? = 3.5 feet. Does this information suggest that the storm is (perhaps temporarily) increasing above the severe rating? Use ? = 0.01.(a) What is the level of significance?1What is the value of the sample test statistic? (Round your answer to two decimal places.)

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