Question

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.

Old Faithful Data Set 23 “Old Faithful” in Appendix B includes data from 250 random eruptions of the Old Faithful geyser. The National Park Service makes predictions of times to the next eruption, and the data set includes the errors (minutes) in those predictions. The accompanying Statdisk display results from using the prediction errors (minutes) to test the claim that the mean prediction error is equal to zero. Comment on the accuracy of the predictions.

Short answer

The hypotheses are as follows.

\(\begin{array}{l}{H_0}:\mu = 0\\{H_1}:\mu \ne 0\end{array}\)

The test statistic is -8.7201, and the p-value is 0.0000.

The null hypothesis is rejected, which implies that there is insufficient evidence to support the claim that the population mean of the prediction error is equal to zero.

This means that the prediction of eruption for the next time is not accurate.

Step-by-step solution

Given information

A sample is taken from the eruption of an old faithful geyser with a sample size of 250 with the claim that the population mean of the prediction error in predicting the eruption is equal to zero.

State the hypotheses

The null hypothesis\({H_0}\)represents the population mean of the prediction error equal to 0. The alternate hypothesis\({H_1}\)represents the population mean of the prediction error, which is not equal to 0.

Let\(\mu \)be the population mean of the prediction error.

State the null and alternate hypotheses.

\(\begin{array}{l}{H_0}:\mu = 0\\{H_1}:\mu \ne 0\end{array}\)

State the test statistic and the p-value

The test statistic and the p-value are represented by the symbols\(t\)and\({\rm{P - value}}\),respectively.

The test statistic and p-value are obtained from the second row and the third row of the given output, respectively. Also, the critical value is obtained from the second row, as stated below.

\(\begin{array}{c}t \approx - 8.7201\\{\rm{P - value}} \approx 0.0000\\{\rm{Critical}}\;t = \pm 1.9695\end{array}\).

State the decision rule

Reject the null hypothesis when the absolute value of the observed test statistics is greater than the critical value. Otherwise, fail to reject the null hypothesis.

\(\begin{array}{c}\left| { - 8.7201} \right| = 8.7201\\ > 1.9695\\t > t\left( {{\rm{critical}}} \right)\end{array}\).

The absolute value of the observed test statistic is significantly larger than the critical value. This implies that there is sufficient evidence to reject the null hypothesis.

Conclusion

As the null hypothesis is rejected, it can be concluded that there is insufficient evidence to support the claim that the population mean of the prediction error is not equal to zero.

As the prediction errors are not equal to 0, the prediction of eruption for the next time is not accurate.