Question

In Exercises 5–20, assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of\({n_1} - 1\)and\({n_2} - 1\).) Is Old Faithful Not Quite So Faithful? Listed below are time intervals (min) between eruptions of the Old Faithful geyser. The “recent” times are within the past few years, and the “past” times are from 1995. Does it appear that the mean time interval has changed? Is the conclusion affected by whether the significance level is 0.05 or 0.01?

Recent	78	91	89	79	57	100	62	87	70	88	82	83	56	81	74	102	61
Past	89	88	97	98	64	85	85	96	87	95	90	95

Short answer

There is sufficient evidence to support the claim that the mean time interval between eruptions has changed at a 0.05 level of significance.

There is insufficient evidence to support the claim that the mean time interval between eruptions has changed at a 0.01 level of significance.

The conclusion of the claim is affected by the change in the significance level.

Step-by-step solution

Given information

The past and recent time intervals (min) between eruptions of the Old Faithful geyser are listed in the table.

Formulation of the hypotheses

Null hypothesis:The past mean time interval between eruptions is equalto the recent mean time interval between eruptions.

\({H_0}\):\({\mu _1} = {\mu _2}\)

Alternative hypothesis:The past mean time interval between eruptions is not equal to the recent mean time interval between eruptions.

\({H_1}\):\({\mu _1} \ne {\mu _2}\)

Calculation of the sample means

The recent mean time interval between eruptions is equal to the following:

\(\begin{array}{c}{{\bar x}_1} = \frac{{\sum\limits_{i = 1}^{{n_1}} {{x_i}} }}{{{n_1}}}\\ = \frac{{78 + 91 + .... + 61}}{{17}}\\ = 78.82\end{array}\)

Therefore, the recent mean time interval between eruptions equals78.82 minutes.

The past mean time interval between eruptions equalsthe following:

\(\begin{array}{c}{{\bar x}_2} = \frac{{\sum\limits_{i = 1}^{{n_2}} {{x_i}} }}{{{n_2}}}\\ = \frac{{89 + 88 + ... + 95}}{{12}}\\ = 89.08\end{array}\)

Therefore, the past mean time interval between eruptions equals89.08 minutes.

Calculation of the sample standard deviations

The standard deviation of the recent time interval between eruptions is computed below:

\(\begin{array}{c}{s_1} = \sqrt {\frac{{\sum\limits_{i = 1}^{{n_1}} {{{({x_i} - {{\bar x}_1})}^2}} }}{{{n_1} - 1}}} \\ = \sqrt {\frac{{{{\left( {78 - 78.82} \right)}^2} + {{\left( {91 - 78.82} \right)}^2} + .... + {{\left( {61 - 78.82} \right)}^2}}}{{17 - 1}}} \\ = 13.97\end{array}\)

Therefore, the standard deviation of the recent time interval between eruptions equals 13.97 minutes.

The standard deviation for the past time interval between eruptions is equal to:

\(\begin{array}{c}{s_2} = \sqrt {\frac{{\sum\limits_{i = 1}^{{n_2}} {{{({x_i} - {{\bar x}_2})}^2}} }}{{{n_2} - 1}}} \\ = \sqrt {\frac{{{{\left( {89 - 89.08} \right)}^2} + {{\left( {88 - 89.08} \right)}^2} + .... + {{\left( {95 - 89.08} \right)}^2}}}{{12 - 1}}} \\ = 9.19\end{array}\)

Therefore, the standard deviation of the past time interval between eruptions is equal to 9.19 minutes.

Calculation of the test statistic

Under the null hypothesis,\({\mu _1} - {\mu _2} = 0\).

The test statistic is computed below:

\(\begin{array}{c}t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}\\ = \frac{{\left( {78.82 - 89.08} \right) - \left( 0 \right)}}{{\sqrt {\frac{{{{\left( {13.97} \right)}^2}}}{{17}} + \frac{{{{\left( {9.19} \right)}^2}}}{{12}}} }}\\ = - 2.385\end{array}\)

Thus, the value of the test statistic is -2.385.

Computation of critical value

Degrees of freedom: The smaller of the two values,\(\left( {{n_1} - 1} \right)\)and\(\left( {{n_2} - 1} \right)\)is considered as the degreesof freedom.

\(\begin{array}{c}\left( {{n_1} - 1} \right) = \left( {17 - 1} \right)\\ = 16\end{array}\)

\(\begin{array}{c}\left( {{n_2} - 1} \right) = \left( {12 - 1} \right)\\ = 11\end{array}\)

The value of the degrees of freedom is the minimum of (16,11) equal to 11.

Now see the t-distribution table for a two-tailed test with a 0.05 level of significance and 11 degrees of freedom.

The critical values are -2.201 and 2.201. The corresponding p-value is equal to 0.0362.

The value of the test statistic does not lie between the values -2.201 and 2.201 and the p-value is less than 0.05. Therefore, the null hypothesis is rejected at a 0.05 significance level.

Conclusion of the test

There is sufficient evidence to support the claim that the mean time interval between eruptions has changed at a 0.05 level of significance.

Changing the level of significance

Let the level of significance be equal to 0.01.

Referring to the t-distribution table, the critical values of t at\(\alpha = 0.01\) with 11 degrees of freedom for a two-tailed test are -3.1058 and 3.1058.

The p-value remains the same and is equal to 0.0362.

The test statistic value equal to -2.385 lies between the two critical values, and the p-value is greater than 0.01.

There is insufficient evidence to support the claim that the mean time interval between eruptions has changed at a 0.01 level of significance.

Thus, the conclusion of the claim changes with the change in the significance level.