Question

hat is the optimal time for a scuba diver to be on the bottom of the ocean? That depends on the depth of the dive. The U.S. Navy has done a lot of research on this topic. The Navy defines the "optimal time" to be the time at each depth for the best balance between length of work period and decompression time after surfacing. Let x = depth of dive in meters, and let y = optimal time in hours. A random sample of divers gave the following data.

x	16.1	26.3	30.2	38.3	51.3	20.5	22.7
y	2.78	2.08	1.48	1.03	0.75	2.38	2.20

(b) Use a 1% level of significance to test the claim that ρ < 0. (Round your answers to two decimal places.)

t	=
critical t	=

Conclusion

9

Fail to reject the null hypothesis. There is insufficient evidence that ρ < 0. Reject the null hypothesis. There is insufficient evidence that ρ < 0.     Fail to reject the null hypothesis. There is sufficient evidence that ρ < 0. Reject the null hypothesis. There is sufficient evidence that ρ < 0.
(c) Find S_e, a, and b. (Round your answers to five decimal places.)

Se	=
a	=
b	=

(d) Find the predicted optimal time in hours for a dive depth of x = 38 meters. (Round your answer to two decimal places.)


(e) Find an 80% confidence interval for y when x = 38 meters. (Round your answers to two decimal places.)

lower limit
upper limit

(f) Use a 1% level of significance to test the claim that β < 0. (Round your answers to two decimal places.)

t	=
critical t	=

Conclusion

(g) Find a 90% confidence interval for β and interpret its meaning. (Round your answers to three decimal places.)

lower limit
upper limit

Answer 1

b)

The hypothesis of testing is H0: pho=0 vs H1:pho<0

x=c(16.1,26.3,30.2,38.3,51.3,20.5,22.7)
y=c(2.78,2.08,1.48,1.03,0.75,2.38,2.2)
cor.test(x,y, alternative="less")
output:
Pearson's product-moment correlation

data: x and y
t = -7.9811, df = 5, p-value = 0.0002492
alternative hypothesis: true correlation is less than 0
95 percent confidence interval:
-1.000000 -0.821737
sample estimates:
cor
-0.962921

form above result we observed that p-value = 0.0002492 is less than 0.01.
Hence we conclude that pho is less than 0

c)

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.962921
R Square	0.927217
Adjusted R Square	0.91266
Standard Error	0.220219
Observations	7
ANOVA
	df	SS	MS	F	Significance F
Regression	1	3.089089	3.089089	63.69731	0.000498
Residual	5	0.242482	0.048496
Total	6	3.331571
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	3.562718	0.234352	15.20242	2.23E-05	2.960297	4.165139	2.960297	4.165139
x	-0.05959	0.007466	-7.98106	0.000498	-0.07878	-0.04039	-0.07878	-0.04039

d) The regression equation is

time=3.562718 - 0.05959* depth

therefore for depth=38, the time is 1.2983

e) 80% CI when x=38

Fit SE Fit 80% CI 80% PI
1.29844 0.105383 (1.14291, 1.45397) (0.938123, 1.65875)

f)

betahat= -0.05959

SE(betahat)=0.007466

t-stat= -7.98106

p-value=0.000498

here p-value is less than 0.01, thence we conclude that beta<0

g) 90% interval for beta

Lower 90.0%	Upper 90.0%
3.090486967	4.034948466
-0.07463057	-0.044542015