Question

The laboratory performed a comparison study of LDL-c values in 40 participants (40 samples are split so are considered matched or pairs) using two different analyzers, the Abbott and the Roche. The Abbott is considered the gold standard method used in many facilities and the Roche is new and of interest due to faster turn-around time and economical factors. The purpose of the comparison study is to determine if there is a significant difference in results by the two methods using correlation, regression and hypothesis testing. It is important to know if the patient results are not significantly different because that will weigh in on the decision to change to the new analyzer or if reported reference ranges would change. We will use 95% confidence limits, a = 0.05, acceptable correlation as 0.98 +0.02, standard error of < 5.0, slope 1.00+ 0.05, intercept 0.0 +5.0 mg/dl. This is based upon the normal LDL-c of 100 mg/dL. Your data analysis tool may provide additional data but it won't be relevant to this type of regression study. So it is important to only evaluate what is asked for below. You are asked to report the a) Pearson correlation b) Regression (R?, slope, intercept, standard error) and c) line plot with labeled x, y and title d) hypothesis testing for difference of means with null hypothesis, alternative hypothesis, critical statistic, calculated p value and t value of this matched continuous data (paired t test) using the provided data below. Remember to validate these calculations based on the two calculated means. If they look close in number, you expect that there is not a statistically significant difference in means based on t and p calculations. Finally, list your e) Discussion based on the linear relationship (values that point to (el) proportional error and (e2) constant error and (€3) standard error), (e4) correlation based on Pearson r and R2 and (e5) hypothesis testing of difference in means to explain if these results indicate a statistical difference in results of the two methods within 95% confidence.

Answer 1

a).

Abbott	Roche
104	103
133	119
115	112
73	78
75	81
82	82
88	86
86	83
75	77
83	86
79	79
79	77
78	78
89	71
74	75
76	78
83	82
82	79
92	93
93	92
79	83
120	123
140	147
141	137
99	97
121	119
87	88
123	121
117	118
101	106
134	128
112	104
92	93
77	81
97	98
101	100
74	82
83	86
155	143
161	162

correlation =

0.976595

b.

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.976595
R Square	0.953738
Adjusted R Square	0.95252
Standard Error	4.956556
Observations	40

	Coefficients
Intercept	7.453012
X Variable 1	0.918006

or equation: Roche = 7.453 + 0.918 Abbott

c.

Roche vs Abbott 180 160 y = 0.918x + 7.453 R2 = 0.9537 140 120 100 Roche 80 60 40 20 0 0 20 40 60 80 100 120 140 160 180 Abbott

d.

t-Test: Paired Two Sample for Means
	Abbott	Roche
Mean	98.825	98.175
Variance	585.5839744	517.4301282
Observations	40	40
Pearson Correlation	0.976594947
Hypothesized Mean Difference	0
df	39
t Stat	0.778647526
P(T<=t) one-tail	0.22044318	not significant
t Critical one-tail	1.684875122
P(T<=t) two-tail	0.440886359	not signficant
t Critical two-tail	2.02269092

The differences are not significant or there is no statistical signifcant difference in Abbott and Roche based on t value and p value.

e.

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.976594947	correlation (strong and positive)
R Square	0.95373769	R-sq (Abbott explains 95.37% variability in Roche.
Adjusted R Square	0.952520261
Standard Error	4.956555994
Observations	40
ANOVA
	df	SS
Regression	1	19246.212
Residual	38	933.5629981
Total	39	20179.775
	Coefficients	Standard Error
Intercept	7.453011513	3.334703359 (proportion error)
X Variable 1	0.918006461	0.032798435 (constant error)

95% CI:

Mean difference -/+ t*SE/sqrt(n) = (98.825 - 98.175) -/+ 0.779 * 4.96/sqrt(40) = 0.65 -/+ 0.61 = (0.04,1.26)

Please rate my answer and comment for doubt.