Question

9. The data below are the hours spent studying and the corresponding test score earned. Assume that the variables x and y have a significant correlation. Hours spent studying, x 0 2 4 5 5 5 6 7 8 Test score, y 4051 64 69 73 75 9390 95 a) Sketch a scatter plot of the data on provided graph paper segment. (2 pts) b) Describe the type of correlation you see if one exists. (1 pts) c) Find the sample correlation coefficient. Include the symbol. (2 pts) d) Perform a hypothesis test to determine if significant linear correlation exits between these two variables at a 5% level of significance. (12 pts) State the hypotheses. Identify the level of significance. Find the standardized test statistic. Include the correct symbol. Indicate the calculator function used. Draw the distribution. Shade the tail(s) and label the values. Make your decision and state why. Write the interpretation and answer the question. e) If there is significant correlation found in the above hypothesis test, write a regression equation for the given information. Include the correct symbols. (3 pts) f) If possible, make a predicted value for the test score a person earned if a person studied 15 hours. Write your answer in a sentence and round to an appropriate number if needed. (2 pts) g) If possible, make a predicted value for the test score a person earned if a person studied 6.5 hours. Write your answer in a sentence and round to an appropriate number if needed. (pts)

Answer 1

a)

Scatterplot - 1 2 3 4 5 6 7 8 9

b)

Linear positive strong relationship

c)

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	42	650	48	2841.6	357.67
mean	4.67	72.22	SSxx	SSyy	SSxy

sample size ,   n =   9          
here, x̅ = Σx / n=   4.67   ,     ȳ = Σy/n =   72.22  
                  
SSxx =    Σ(x-x̅)² =    48.0000          
SSxy=   Σ(x-x̅)(y-ȳ) =   357.7          
                  
estimated slope , ß1 = SSxy/SSxx =   357.7   /   48.000   =   7.4514
                  
intercept,   ß0 = y̅-ß1* x̄ =   37.4491          
                  
so, regression line is   Ŷ =   37.4491   +   7.4514   *x
                  
SSE=   (SSxx * SSyy - SS²xy)/SSxx =    176.442          
                  
std error ,Se =    √(SSE/(n-2)) =    5.021          
                  
correlation coefficient ,    r = Sxy/√(Sx.Sy) =   0.9685          

d)

Ho:   ρ = 0       tail=   2  
Ha:   ρ ╪ 0              
n=   9              
alpha,α =    0.05              
correlation , r=   0.9685              
t-test statistic = r*√(n-2)/√(1-r²) =        10.283          
DF=n-2 =   7              
p-value =    0.0000             
Decison:   p value < α , So, Reject Ho             

There is significant relatioship

e)

  
so, regression line is   Ŷ =   37.4491   +   7.4514   *x

f)

Predicted Y at X=   15   is                  
Ŷ =   37.44907   +   7.451389   *   15   =   149.220

g)

Predicted Y at X=   6.5   is                  
Ŷ =   37.44907   +   7.451389   *   6.5   =   85.883

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