Glomerular filtration rate (GFR) is the most important parameter of renal function assessed in renal transplant recipients. Although inulin clearance is regarded as the gold standard measure of GFR, its use in clinical practice is limited. Krieser et al. examined the relationship between the inverse of Cystatin C (a cationic basic protein measured in mg/L) and inulin GFR as measured by technetium radionuclide labeled diethylenetriamine penta-acetic acid) (DTPA GFR) clearance (ml/min/1.73 m2). The results of 27 tests are shown in the following table:   DTPA GFR          1/Cystatin C DTPA GFR              1/Cystatin C 18                            0.213 21                            0.265 21                            0.446 23                            0.203 27                            0.369 27                            0.568 30                            0.382 32                            0.383 32                            0.274 32                            0.424 36                            0.308 37                            0.498 41                            0.398 42                                0.485 42                                0.427 43                                0.562 43                                0.463 48                                0.549 48                                0.538 51                                0.571 55                                0.546 58                                0.402 60                                0.592 62                                0.541 67                                0.568 68                                0.800 68                                0.667   Use the data to solve the following: 4. Test if there is a significant relationship between variables at alpha = 0.05 (show 6 steps). 5. Compute for the regression model and predict the cystatin c if DTPA GFR is 50. (Use two decimal places).

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[four&five] Instructions: Kindly answer the needed answers below. Examples on how to answer are already included in the photo. Show solution.

Glomerular filtration rate (GFR) is the most important parameter of renal function assessed in renal transplant recipients. Although inulin clearance is regarded as the gold standard measure of GFR, its use in clinical practice is limited. Krieser et al. examined the relationship between the inverse of Cystatin C (a cationic basic protein measured in mg/L) and inulin GFR as measured by technetium radionuclide labeled diethylenetriamine penta-acetic acid) (DTPA GFR) clearance (ml/min/1.73 m2). The results of 27 tests are shown in the following table:

 

DTPA GFR          1/Cystatin C

DTPA GFR              1/Cystatin C

18                            0.213

21                            0.265

21                            0.446

23                            0.203

27                            0.369

27                            0.568

30                            0.382

32                            0.383

32                            0.274

32                            0.424

36                            0.308

37                            0.498

41                            0.398

42                                0.485

42                                0.427

43                                0.562

43                                0.463

48                                0.549

48                                0.538

51                                0.571

55                                0.546

58                                0.402

60                                0.592

62                                0.541

67                                0.568

68                                0.800

68                                0.667

 

Use the data to solve the following:

4. Test if there is a significant relationship between variables at alpha = 0.05 (show 6 steps).

5. Compute for the regression model and predict the cystatin c if DTPA GFR is 50. (Use two decimal places).

Step 4: Computation
- 0.93.
1-
L- 0.93.
L- 0.93 V33. 23
I- 10.73
Step 5: Dasislan
Since the 10. 73 > 2 16, reject Null Hypothesis (Ho) and Accept Aiternative
Hypothesis (Ha).
Step 6: Conolucion
Therefore, there is a significant relationship betwaen heights and waights of the
20 colloge students.
e Compute for the regression model and predict the cystatin e if DTPA GFR is 50.
(Use hwo decimai pieces).
Nest, we use these values to find the value ot
)-
2(2) - RSLS
4.IN
We then find the value ofb
LET - (
1578 - 26145
20
The regression aquation is -755r+(-54. 7) or -755r - s4 n.
We con now use the regression equatfon to estimote the weight of the students given the
height. For instance. I the height of the students is 171 cm. then the estimated weight s
=0 755- 54.97
- 0.755(171) - 5497
- 74. 41
The opproximate weight is 74.41 ko
Transcribed Image Text:Step 4: Computation - 0.93. 1- L- 0.93. L- 0.93 V33. 23 I- 10.73 Step 5: Dasislan Since the 10. 73 > 2 16, reject Null Hypothesis (Ho) and Accept Aiternative Hypothesis (Ha). Step 6: Conolucion Therefore, there is a significant relationship betwaen heights and waights of the 20 colloge students. e Compute for the regression model and predict the cystatin e if DTPA GFR is 50. (Use hwo decimai pieces). Nest, we use these values to find the value ot )- 2(2) - RSLS 4.IN We then find the value ofb LET - ( 1578 - 26145 20 The regression aquation is -755r+(-54. 7) or -755r - s4 n. We con now use the regression equatfon to estimote the weight of the students given the height. For instance. I the height of the students is 171 cm. then the estimated weight s =0 755- 54.97 - 0.755(171) - 5497 - 74. 41 The opproximate weight is 74.41 ko
The following are the format (examples already provided) for ancwering the data. Don't
forget to put a label (e.0. "atep 1" or the letter a"). ahow colution
a Construct a scatter-plot
Exomple: A study of the amount of raintaland the quality of or oollution removed
produced the foilowing dota:
Doly Ronfot 4.3. 45. 9.5A. 6.1. 5.2. 38. 21,75
Paticulote Removed: 26. 21. 16. 18. 14. 18. 32. 41.8
Anwer: The study shows thot the dely rointal
is negative corelction with the porticulcte removec
b. compute the Pearson rand interpret (shaw table).
PORMULA
c. compute for the coeficient of determination and interpret.
Carrelation coeficients are interpreted in a relative sort of way by talking about
high, moderate, or low. However, the best way to evaluate an ris by the use of
the coefficient of determination, denoted by.
In the given example, the computed r= 0.93. The coefficient of determination,
- 0.86. This means that for the 20 college students in the sampie 86% of the
variation weighs could be attributed to the varlation in heights. The rest (14%) is
chance variation.
a Test there is a significant relationship betwoen varlabies at alpha = 0.05 (show
and fallow 6 steps).
Step 1: Identitv He and H.
Ha: (P=0) There is no significant relationship between haights and weights of 20
college students.
Ha: (P-0) There is a significant difference batween haights and weights of 20
college students.
Step 2: Lavel of Blanifioanos
a-0. 05
af =n-2
= 20 -2
= 18
Critical value t 2.10 (based on the t-lable)
Step 3: Iastof statietta
Transcribed Image Text:The following are the format (examples already provided) for ancwering the data. Don't forget to put a label (e.0. "atep 1" or the letter a"). ahow colution a Construct a scatter-plot Exomple: A study of the amount of raintaland the quality of or oollution removed produced the foilowing dota: Doly Ronfot 4.3. 45. 9.5A. 6.1. 5.2. 38. 21,75 Paticulote Removed: 26. 21. 16. 18. 14. 18. 32. 41.8 Anwer: The study shows thot the dely rointal is negative corelction with the porticulcte removec b. compute the Pearson rand interpret (shaw table). PORMULA c. compute for the coeficient of determination and interpret. Carrelation coeficients are interpreted in a relative sort of way by talking about high, moderate, or low. However, the best way to evaluate an ris by the use of the coefficient of determination, denoted by. In the given example, the computed r= 0.93. The coefficient of determination, - 0.86. This means that for the 20 college students in the sampie 86% of the variation weighs could be attributed to the varlation in heights. The rest (14%) is chance variation. a Test there is a significant relationship betwoen varlabies at alpha = 0.05 (show and fallow 6 steps). Step 1: Identitv He and H. Ha: (P=0) There is no significant relationship between haights and weights of 20 college students. Ha: (P-0) There is a significant difference batween haights and weights of 20 college students. Step 2: Lavel of Blanifioanos a-0. 05 af =n-2 = 20 -2 = 18 Critical value t 2.10 (based on the t-lable) Step 3: Iastof statietta
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