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Technology. In Exercises 9–12, test the given claim by using the display provided from technology. Use a 0.05 significance level. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Tornadoes Data Set 22 “Tornadoes” in Appendix B includes data from 500 random tornadoes. The accompanying StatCrunch display results from using the tornado lengths (miles) to test the claim that the mean tornado length is greater than 2.5 miles.

Short Answer

Expert verified

The hypotheses are as follows.

\[\begin{array}{l}{H_0}:\mu = 2.5\\{H_1}:\mu > 2.5\end{array}\]

The test statistic is 0.7586.

The p-value is 0.2242.

The null hypothesis is failed to be rejected.

Thus, it is concluded that there is not sufficient evidence to support the claim that the tornado length is greater than 2.5 miles.

Step by step solution

01

Given information

A sample is taken from the tornado length with a sample size of 500 with the claim that the population mean of the tornado length is greater than \[2.5\,\,{\rm{miles}}\].

02

State the hypotheses

The null hypothesis\[{H_0}\]represents the population mean of the tornado length, which is equal to\[2.5\,\,{\rm{miles}}\]. As equality is not present in the claim, the alternate hypothesis\[{H_1}\]represents the population mean of the tornado length, which is greater than\[2.5\,\,{\rm{miles}}\].

Let\[\mu \]be the population mean of the tornado length in miles.

State the null and alternate hypotheses.

\[\begin{array}{l}{H_0}:\mu = 2.5\\{H_1}:\mu > 2.5\end{array}\].

03

State the test statistic and the p-value from the given output

State the test statistic (T-stat) and the p-value obtained from the fifth column and the sixth column of the given output, respectively.

\[\begin{array}{c}t = 0.75865215\\ \approx 0.7586\\p = 0.2242\\ \approx 0.2242\end{array}\]

04

State the decision rule for the p-value

Reject the null hypothesis when the p-value is less than the significance level. Otherwise, fail to reject the null hypothesis.

Suppose the significance level is equal to\[0.05\].

The p-value is greater than . Thus, the null hypothesis is failed to be rejected at a 0.05 level of significance.

05

Conclusion

As the null hypothesis is failed to be rejected, it can be concluded that there is not sufficient evidence to support the claim that the population mean of the tornado length is greater than \[2.5\,\,{\rm{miles}}\].

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Most popular questions from this chapter

Finding Critical t Values When finding critical values, we often need significance levels other than those available in Table A-3. Some computer programs approximate critical t values by calculating t=df×eA2/df-1where df = n-1, e = 2.718, A=z8×df+3/8×df+1, and z is the critical z score. Use this approximation to find the critical t score for Exercise 12 “Tornadoes,” using a significance level of 0.05. Compare the results to the critical t score of 1.648 found from technology. Does this approximation appear to work reasonably well?

In Exercises 9–12, refer to the exercise identified. Make subjective estimates to decide whether results are significantly low or significantly high, then state a conclusion about the original claim. For example, if the claim is that a coin favours heads and sample results consist of 11 heads in 20 flips, conclude that there is not sufficient evidence to support the claim that the coin favours heads (because it is easy to get 11 heads in 20 flips by chance with a fair coin).

Exercise 8 “Pulse Rates”

Critical Values. In Exercises 21–24, refer to the information in the given exercise and do the following.

a. Find the critical value(s).

b. Using a significance level of = 0.05, should we reject H0or should we fail to reject H0?

Exercise 18

In Exercises 13–16, refer to the exercise identified and find the value of the test statistic. (Refer to Table 8-2 on page 362 to select the correct expression for evaluating the test statistic.)

Exercise 6 “Cell Phone”

Test Statistics. In Exercises 13–16, refer to the exercise identified and find the value of the test statistic. (Refer to Table 8-2 on page 362 to select the correct expression for evaluating the test statistic.)

Exercise 7 “Pulse Rates”

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