Question

A researcher studies water darity at the same location in a lake on the same dates during the course of a year and repeats the measurements on the same dates 5 years later. The researcher immerses a weighted disk painted black and white and measures the depth (in inches) at which it is no longer visible. The collected data is given in the table below. Complete parts (a) through (c) below. Observation 1 2 3 4 5 Date 1/253/195/30 7/3 9/13 1117 Initial Depth, X 402 50.1 46.3 744 504 64.1 Depth Five Years Later, Y 48.4 48.3 49.B 71.7 54.9 60.6 Click the icon to view the table of critical t-values Why is it important to take the measurements on the same date? O A. Using the same dates makes it easier to remember to take samples. OB. Using the same dates maximizes the difference in water clarity OC. Using the same dates makes the second sample dependent on the first and reduces variability in water clarity attributable to date. O D. Those are the same dates that all biologists use to take water clarity samples b) Does the evidence suggest that the clarity of the link in improving at the a = 0.05 level of significance? Note that the normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outlers. Let - X-Y, Identify the null and alternative hypotheses HoHo v V Click to select your answers).

Answer 1

a)

OB. Using the same dates maximizes the difference in water clarity

Ho :   µd=   0                  
Ha :   µd <   0                 
                          
Level of Significance ,    α =    0.05                  
                          
sample size ,    n =    6                  
                          
mean of sample 1,    x̅1=   54.250                  
                          
mean of sample 2,    x̅2=   55.283                  
                          
mean of difference ,    D̅ =ΣDi / n =   -1.0333                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    4.1788                  
                          
std error , SE = Sd / √n =    4.1788   / √   6   =   1.7060      
                          
t-statistic = (D̅ - µd)/SE = (   -1.033333333   -   0   ) /    1.7060   =   -0.61
                          
Degree of freedom, DF=   n - 1 =    5                  
t-critical value , t* =        -2.02 [excel function: =t.inv(α,df) ]            
                          
Do not reject null hypothesis     

There is not sufficient evidence that  

OD. -6 42 Differences 2

It supports the above part because the difference is symmetric

Please let me know in case of any doubt.

Thanks in advance!

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