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