Question

) Bicarbonate in replicate samples of horse blood was measured four times by each of the two methods with the following results? Method 1: 51.40, 51.24, 51.18, 51.43 mM Method 2: 50.70, 49.49, 50.01, 50.15 mM a) Find the mean, standard deviation, percent relative standard deviation (%RSD) for each method. b) Are the standard deviations significantly different at 90% confidence level? c) Are the mean values significantly different at 90% confidence level? d) Using Grubbs test decide if number 49.49 should be retained in method 2 at 95% confidence level.

Answer 1

A. The mean, standard deviation, and percent relative standard deviation are:

	M1	M2
	51.4	50.7
	51.24	49.49
	51.18	50.01
	51.43	50.15
Mean	51.3125	50.0875
Standard deviation	0.121484	0.497351
%RSD	0.24%	0.99%

Mean is calculated using =average() ,Excel formula,Standard deviation =stdev.s() and relative standard deviation = (standard deviation/mean)*100

b.Standard deviation significantly different

Null Hypothesis,H0:  

Alternative Hypothesis,Ha:

We have

F = 0.494^2/0.1215^2=0.2474/0.014758 = 16.76

Fcritical value = 9.2766

As the F-value is greater than F-critical value,therefore we will reject the null and conclude that there is a significant difference between the standard deviation

c.We have

H0:m1=m2

Ha:m1m2

Significance level,alpha=0.1

Pooled variance = (3*0.1215^2+3*0.4974^2)/6 = 0.1311

Standard error = sqrt(0.13111*(1/4+1/4)) = 0.256

Difference in mean = 51.3125-50.0875 = 1.2250

t= 1.2250/0.256 = 4.785

p-value = 0.003

As the p-value is less than the significance level,therefore we will reject the null at 90% level of confidence.

d.We have Gexp = absolute value of (49.49- Average)/standard deviation = |49.49-50.0875|/0.497351 = 1.2014

For 95% confidence at N=4,we have Gcritical = 1.46

As the G calculated value is less than the G critical value ,therefore the number should be retained.