91影视

A sample of 500 tornadoes is considered. It is claimed that the mean tornado length is greater than 2.5 miles.

Since $蟽$is unknown, the test statistic used to test the given claim is the t-score which follows Student鈥檚 t distribution with $\left(n-1\right)$ degrees of freedom.

Let n be the sample size.

The formula to compute the value of the critical score is given below:

$t=\sqrt{df脳\left(e^{A^2/df}-1\right)}$

Where,

dfis the degree of freedom which is equal to

e is equal to 2.718

The formula to compute A is given below:

$A=\frac{z\left(8\left(df\right)+3\right)}{\left(8\left(df\right)+1\right)}$

Here, n=500.

Thus, the value of the degrees of freedom is computed below:

$$\begin{gathered} df=n-1 \\ =500-1 \\ =499 \end{gathered}$$

Let the level of significance be equal to $伪=0.05$

Since the claim involves a greater than sign, the test would be right-tailed.

The value of the z-score for; $伪=0.05$ a right-tailed test is equal to 1.645.

Thus, the value of A is computed below:

$$\begin{gathered} A=\frac{z\left(8\left(df\right)+3\right)}{\left(8\left(df\right)+1\right)} \\ =\frac{1.645\left(8\left(499\right)+3\right)}{\left(8\left(499\right)+1\right)} \\ =1.646 \end{gathered}$$

The critical score using the formula stated above is computed as follows:

$$\begin{gathered} t=\sqrt{df脳\left(e^{A^2/df}-1\right)} \\ =\sqrt{df脳\left(e^{\frac{A^2}{df}}-1\right)} \\ =\sqrt{499脳\left(e^{\frac{{\left(1.646\right)}^2}{499}}-1\right)} \\ =1.648 \end{gathered}$$

Thus, the value of the critical score is equal to 1.648.

The value of the critical score at $伪=0.05$ using the formula stated in the question is equal to 1.648.

The value of the critical score at $伪=0.05$ using the t-distribution table/technology is equal to 1.648.

Since the two critical values using the two methods are approximately equal, the given formula to approximate the critical score works considerably well.