Question

In Exercises 5–16, test the given claim.

Old Faithful Listed below are time intervals (min) between eruptions of the Old Faithfulgeyser. The “recent” times are within the past few years, and the “past” times are from 1995.

Does it appear that the variation of the times between eruptions has changed?

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 enough evidence to support the claim that the variation of the times between eruptions has changed

Step-by-step solution

Given information

Two samples are considered. One sample represents the recent time intervals between eruptions with a sample size equal to 17,and the other represents past time intervals between eruptions with a sample size equal to 12.

It is claimed that the variation inthe recent eruption times is not equal to the variation in the pasteruption times.

Hypotheses

Let\({\sigma _1}\)and\({\sigma _2}\)be the population standard deviationsof the recent eruption timesandthe past eruption timesrespectively.

Null hypothesis: The population standard deviation of the recent eruption timesis equal to the population standard deviation of the past eruption times.

Symbolically,

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

Alternate Hypothesis: The population standard deviation of the recent eruption times is not equal to the population standard deviation of the past eruption times.

Symbolically,

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

Sample mean, sample size, and sample standard deviations  

The sample variance has the following formula:

\({s^2} = \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {x - \bar x} \right)}^2}} \)

The sample mean recent eruption time is equal to:

\(\begin{array}{c}{{\bar x}_1} = \frac{{78 + 91 + ....... + 61}}{{17}}\\ = 78.82\end{array}\)

The sample standard deviation of the recent eruption times is computed below:

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

Therefore, the standard deviation of the recent time interval between eruptions is equal to 13.97 minutes.

The sample mean past eruption time is equal to:

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

The sample standard deviation of the past eruption times is computed below:

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

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

Compute the test statistic

Since two independent samples involve a claim about the population standard deviation, apply an F-test.

Consider the larger sample variance to be\(s_1^2\)and the corresponding sample size to be\({n_1}\).

The following values are obtained:

\({\left( {13.97} \right)^2} = 195.1609\)

\({\left( {9.19} \right)^2} = 84.4561\)

Here,\(s_1^2\)is the sample variance corresponding to recent eruption times and has a value equal to 195.1609 minutes squared.

\(s_2^2\)is the sample variance corresponding to the past eruption times and has a value equal to 84.4561 minutes squared.

Substitute the respective values to calculate the F statistic:

\(\begin{array}{c}F = \frac{{s_1^2}}{{s_2^2}}\\ = \frac{{{{\left( {13.97} \right)}^2}}}{{{{\left( {9.19} \right)}^2}}}\\ = 2.311\end{array}\)

State the critical value and the p-value

The value of the numerator degrees of freedom is equal to:

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

The value of the denominator degrees of freedom is equal to:

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

For the F test, the critical value corresponding to the right-tail is considered.

The critical value can be obtained using the F-distribution table with numerator degrees of freedom equal to 16and denominator degrees of freedom equal to 11 for a right-tailed test.

The level of significance is equal to:

\(\begin{array}{c}\frac{\alpha }{2} = \frac{{0.05}}{2}\\ = 0.025\end{array}\)

Thus, the critical value is equal to 3.3044.

The two-tailed p-value for F equal to 2.311 is equal to 0.1631.

Conclusion

Since the test statistic value is less than the critical value and the p-value is greater than 0.05, the null hypothesis fails to be rejected.

Thus, there is enough evidence to supportthe claimthat the variation of the times between eruptions has changed