Question

Some geysers such as Old Faithful in Yellowstone National Park are remarkably consistent in the periodicity of their eruption. For example, in 1988, 6,900 timed intervals between eruptions for Old Faithful averaged 76.17 minutes, with the shortest observed interval 41 minutes and the longest 114 minutes. In the past 120 years Old Faithful's yearly average interval has always been between 60 and 79 minutes.

It is also well known that the relationship between the length of the eruption and the length of the subsequent interval duration is a positive one. Suppose the following data were collected over a several day period.

1: Compute r, the Pearson correlation coefficient

2: At the 0.05 level of significance, test the null hypothesis that the (“eruption time” and the “interval duration”) population correlation coefficient [ρ] is equal to 0.

3: Compute and use the regression equation you came up with in the previous part (namely “f”) to predict the “interval duration” for an “eruption time” of 6 minutes.

Observation	Eruption time (min)	Interval duration (min)
1	1.5	50
2	2.1	56
3	2.4	65
4	3.2	71
5	2.9	70
6	2.5	66
7	2.2	57
8	3.5	76
9	3.0	69
10	3.5	76
11	4.1	82
12	2.0	57
13	4.6	89
14	2.8	70
15	5.0	95
16	3.6	75
17	4.0	80
18	2.4	67
19	3.5	77
20	4.9	94

Answer 1

	X	Y
	1.5	50
	2.1	56
	2.4	65
	3.2	71
	2.9	70
	2.5	66
	2.2	57
	3.5	76
	3.0	69
	3.5	76
	4.1	82
	2.0	57
	4.6	89
	2.8	70
	5.0	95
	3.6	75
	4.0	80
	2.4	67
	3.5	77
	4.9	94
MEAN	3.1850	72.1000
S.D	0.9635	11.9369
r	0.9866

*************************************************************************

(b) We frame the Null Hypothesis

(Or)

The variables are uncorrelated in the population.

To test ; we use the " t - Statistic "

Since = 0.05 = 5%

; So, we Reject at 5% Level of Significant.

Therefore we conclude that "The variables are correlated in the population".

***********************************************************************************************

(c)

To predict the “interval duration” for an “eruption time” of 6 minutes we proceed as follows.

Given X = 6; so we have to find out the value of Y.

For this we have to fit the Regression Equation of Y on X

The Regression Equation of Y on X is

Substitute X = 6 in the above fitted Regression Equation

***************************************************************

NOTE: The critical value of t has been extracted from the tabulated value which posted below.