The data show the time intervals after an eruption to the ne

The data show the time intervals after an eruption (to the next eruption) of a certain geyser. Find the regression equation, letting the first variable be the independent (x) variable. Find the best predicted time of the interval after an eruption given that the current eruption has a height of 92 feet. Use a significance level of 0.05.

Height (ft)

70

73

104

122

74

107

76

97

Interval after (min)

69

62

82

91

71

84

68

75

What is the regression equation?

y= [__]+[__]x

(Round to two decimal places as needed.)

What is the best predicted value?

y[__] minutes (Round to one decimal place as needed.)

Icon below

Critical Values of the Pearson Correlation Coefficient r

H0:

rhoequals=0

H1:

rhonot equals0,

H0

n

alphaequals=0.05

alphaequals=0.01

4

0.950

0.990

5

0.878

0.959

6

0.811

0.917

7

0.754

0.875

8

0.707

0.834

9

0.666

0.798

10

0.632

0.765

11

0.602

0.735

12

0.576

0.708

13

0.553

0.684

14

0.532

0.661

15

0.514

0.641

16

0.497

0.623

17

0.482

0.606

18

0.468

0.590

19

0.456

0.575

20

0.444

0.561

25

0.396

0.505

30

0.361

0.463

35

0.335

0.430

40

0.312

0.402

45

0.294

0.378

50

0.279

0.361

60

0.254

0.330

70

0.236

0.305

80

0.220

0.286

90

0.207

0.269

100

0.196

0.256

Height (ft)	70	73	104	122	74	107	76	97
Interval after (min)	69	62	82	91	71	84	68	75

Solution

a)y = 32.866+ 0.469*( Height)......

height = 92..predicted time = 32.866+ (0.469*92) = 76.014......

b) corelation coefficient = cov( height , predicted time) / [ s.d( height) * s.d( predicted time) ]

= 0.951...

s_r = sqrt [ ( 1 - r^2) / (n-2) ] = 0.154.......

test statistic = r / s_r = 0.951 / 0.154 = 6.175= absolute value...d.f = 8-2 = 6..
p-value = 0.0008..

critical value= 0.950 ( for alpha = 0.05 and d.f = 4)..

so, absolute value > criticalvalue!

so, reject H0 in favour of H1....