Question

What 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. 15.1 22.3 29.2 38.3 51.3 20.5 22.7 y 2.58 1.98 1.88 1.03 0.75 2.38 2.20 (a) Find Ex, Xy, Ex2, Ey?, Exy, and r. (Round r to three decimal places.) Ey = Ex2 Sy2 = (b) Use a 1% level of significance to test the claim that p < 0. (Round your answers to two decimal places.) critical t = Conclusion Fail to reject the null hypothesis. There is sufficient evidence that p < 0. Reject the null hypothesis. There is insufficient evidence that p < 0. Fail to reject the null hypothesis. There is insufficient evidence that p < 0. Reject the null hypothesis. There is sufficient evidence that p < 0. (c) Find Sera, and b. (Round your answers to four decimal places.) Se = a = b = (d) Find the predicted optimal time in hours for a dive depth of x = 37 meters. (Round your answer to two decimal places.) hr (e) Find an 80% confidence interval for y when x = 37 meters. (Round your answers to two decimal places.) lower limit hr upper limit hr (0) Use a 1% level significance to test the claim that B < 0. (Round your answers to two decimal places.) critical Conclusion Fail to reject the null hypothesis. There is insufficient evidence that B < 0. Reject the null hypothesis. There is sufficient evidence that B < 0. Fail to reject the null hypothesis. There is sufficient evidence that ß < 0. Reject the null hypothesis. There is insufficient evidence that B < 0.

Answer 1

a)

	X	Y	XY	X²	Y²
total sum	199.400	12.800	314.66	6612.060	26

sample size ,   n =   7          
here, x̅ = Σx / n=   28.49   ,     ȳ = Σy/n =   1.83  
                  
SSxx =    Σ(x-x̅)² =    932.0086          
SSxy=   Σ(x-x̅)(y-ȳ) =   -50.0          
                  
estimated slope , ß1 = SSxy/SSxx =   -50.0   /   932.009   =   -0.0536
                  
intercept,   ß0 = y̅-ß1* x̄ =   3.3554          
                  
so, regression line is   Ŷ =   3.3554   +   -0.0536   *x
                  
SSE=   (SSxx * SSyy - SS²xy)/SSxx =    0.156          
                  
std error ,Se =    √(SSE/(n-2)) =    0.176          
                  
correlation coefficient ,    r = Sxy/√(Sx.Sy) =   -0.972   

b)

Ho:   ρ = 0       tail=   2  
Ha:   ρ ╪ 0              
n=   7              
alpha,α =    0.01              
correlation , r=   -0.9721              
t-test statistic = r*√(n-2)/√(1-r²) =        -9.27   
DF=n-2 =   5              
p-value =    0.0001   
Decison:   p value < α , So, Reject Ho              
critical t-value =    -3.36
  

Reject the null hypothesis. There is sufficient evidence that p < 0.

c)

std error ,Se =    √(SSE/(n-2)) =    0.176

a = intercept,   ß0 = y̅-ß1* x̄ =   3.3554

b =estimated slope , ß1 = SSxy/SSxx =   -50.0   /   932.009   =   -0.0536

d)

Predicted Y at X=   37   is                  
Ŷ =   3.35539   +   -0.053599   *   37   =   1.372

e)

standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    0.083              
margin of error,E=t*Std error=t* S(ŷ) =   1.4759   *   0.0829   =   0.1223
                  
Confidence Lower Limit=Ŷ +E =    1.372   -   0.122   =   1.250
Confidence Upper Limit=Ŷ +E =   1.372   +   0.122   =   1.495

Thanks in advance!

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