Chapter 4 – Statistics
1. Consider the following data.
45.5 44.2 59.4 55.2 49.6 47.3 46.7
Find the (a) mean (b) standard deviation and (c) variance
Mean x = 49.7
St. Dev s = 5.6
Variance s2 = 31.2
2. Suppose a Gaussian population has a mean of 2000 and a standard deviation of 75. What fraction of the population lies in the following interval?
(a) > 2000 (b) 1925 – 2075 (c) 1775 (d) 1850 – 2150 (e) 1898 – 1958
a) z = (2000 -2000)/75 = 0 P(0 - ∞) = 0.5000
50.00 %
b) z = (1975 - 2000)/75 = -1 P(-1 to 0 ) = P (0 to 1) = 0.3413
z = (2075 - 2000)/75 = 1 P (0 to 1) = 0.3413
P(-1 to 1) = P(-1 to 0 ) + P (0 to 1) = 0.3413 + 0.3413 = 0.6826
69.26 %
c) z = (1775 – 2000)/75 = -3 P(-3 to 0) = P( 0 to 3) = 0.4987
P(3 to ∞) = P(0 to ∞) - P( 0 to 3) = 0.5000 – 0.4987 = 0.0013
0.13 %
d) z = (1850 -2000)/75 = -2 P(-2 to 0) = P(0 to 2) = 0.4773
z = (2150 -2000)/75 = 2 P(0 to 2) = 0.4773
P(-2 to 2) = P(-2 to 0) + P(0 to 2) = 0.4773 + 0.4773 = 0.9546
95.46 %
e) z = (1898 -2000)/75 = -1.4 P(-1.4 to 0) = P(0 to 1.4) = 0.4192
z = (1958 -2000)/75 = -0.6 P(-0.6 to 0) = P(0 to-0.6) = 0.2258
P(-1.4 to -0.6) = P(-1.4 to 0) - P(-0.6 to 0) = 0.4192 + 0.2258 = 0.1934
19.34 %
3. Two methods were used to measure the specific activity (units of enzyme activity per milligram of protein) of an enzyme. One unit of enzyme activity is defined as the amount of enzyme that catalyzes the formation of one micromole of product per minute under specified conditions.
Enzyme activity (five replications)
Method 1: 139 147 160 158 135
Method 2: 148 159 156 164 159
Is the mean value of method 1 significantly different from the mean value of method 2 at the 95% confidence level?
Let’s assume that the measurements were made on the same sample by the two different testing methods; therefore, we use the comparison of individual differences test.
Trial di |di – d| (di –d)2
1 -9 0.4 0.16
2 -12 2.6 6.76
3 4 13.4 179.56
4 -6 3.4 11.56
5 -24 14.6 213.16
S 411.2
Mean d = -9.4
St. Dev. s = (411.2/5-1)0.5 = 10.14
tcalc = |-9.4|(5)0.5/10.14 = 2.07
ttab for n = 5, DF = 4, at 95% confidence is 2.776
ttab > tcal ; therefore, there is no significant difference.
4. The Ti content (wt %) of two different ore samples was measured several times by the same method. Are the mean values significantly different at the 95% confidence level?
Sample 1: 0.013 4 0.013 8 0.012 8 0.013 3 0.013 7
Sample 2: 0.013 5 0.014 2 0.013 7 0.014 1 0.014 3
Here we have two different samples measured using the same method.
Sample 1 x1 = 0.0134 s1 = 3.9 *10-4
Sample 2 x2 = 0.0140 s2 = 3.4 *10-4
sp = {[s12(n1-1) + s22(n2-1)]/(n1 + n2 -2)}0.5 =
= {[(3.9 *10-4)2(5 -1) + (3.4 *10-4)2(5 -1)]/(5 + 5 -2)} 0.5
= {1.07 *10-6/ 8}0.5
= 3.7*10-4
tcalc = |x1- x2| [n1n2/(n1+n2)]0.5/sp
= |0.0134 – 0.0140|[5*5/(5+5)]0.5/3.7*10-4
= (6*10-4)(1.58)/ 3.7*10-4 = 2.56
ttab at 95 % confidence for 8 degrees of freedom is 2.306;
therefore, tcalc > ttab => the samples are statistically significantly different.
5. Using the Q test, decide whether the value 1.459 should be rejected from the set of results: 1.422, 1.429, 1.459, 1.426, 1.426, 1.428.
1.422 1.426 1.426 1.428 1.429 1.459
Gap = 1.459 – 1.429 = 0.030
Range = 1.459 -1.422 = 0.037
Qcalc = Gap/Range = 0.030/0.037 = 0.81
Qtab for 6 measurements at 90 % confidence is 0.56.
Since Qcalc > Qtab, the data point can be rejected.
6. A set of absorbance measurements were obtained as 0.985, 0.979, 0.961, 0.996 and 0.978 AU. What are the 50%, 95% and 99% confidence limits of there data.
Mean x = 0.980
St. dev s = 1.27*10-2
n=5 DF = 4
50 % C.L.
t = 0.741
m = x ± ts/n0.5
= 0.980 ± 0.741*1.27*10-2/40.5
= 0.980 ± 4.71*10-3 = 0.975 … 0.985
95 % C.L.
t = 2.776
m = x ± ts/n0.5
= 0.980 ± 2.776*1.27*10-2/40.5
= 0.980 ± 1.76*10-2 = 0.962 … 0.998
99 % C.L.
t = 4.604
m = x ± ts/n0.5
= 0.980 ± 4.604*1.27*10-2/40.5
= 0.980 ± 2.92*10-2 = 0.951 … 1.009