1A. A sample of lard was isolated from a Super-Jumbo order of Burger Emporium fr
ID: 883680 • Letter: 1
Question
1A.
A sample of lard was isolated from a Super-Jumbo order of Burger Emporium french fries. The mass was separately determined five times as 30, 30, 52, 34 and 32 mg. Calculate the Q value for the highest measurement. Round your answer correctly to 2 decimal places, include the leading zeros.
b.True/False: The 52 mg measurement is statistically different than the others at the 90% confidence level.
c.Keeping or rejecting the 52 mg measurement (depending on the results of the Q-test), calculate the mean of the modified data set. (Report answer to 1 decimal place and with NO units.)
d.Calculate the estimated standard deviation of the modified data set. (Report to 1 decimal place and with NO units.)
Explanation / Answer
Dixon's Q-test for detection of a single outlier in a data set.
The test is applied as follows:
(i) The N values comprising the set of observations under examination are arranged in ascending order:
X1 < X2 < …… XN-1 < XN
(ii) The statistic experimental Q-value (Qexp) is calculated as follows.
It is a ratio defined as the difference of the suspect value from its nearest one divided by the range of the values (Q = rejection quotient). Thus, for testing XN (as possible outlier) we use the following equation to calculate Qexp:
Qexp = XN – XN-1 / XN – X1
(iii) The obtained Qexp value is then compared to the critical Q-value (Qcrit) or Q-test value (Qtest) found in tables. This critical/test value should correspond to the confidence level (CL) we have decided to run the test (usually: CL=95%).
Answer for a)
As per the above notes the given mass values are arranged in increasing order as shown below:
30, 30, 32, 34, and 52
Qvalue for 52 = 52-34 / 52-30 = 0.82
Answer for b) True
The 52 mg measurement is statistically different than the others at the 90% confidence level as calculated value (0.82) for 52 mg measurement is higher than critical Q-value (Qcrit) or Q-test (Qtest) found in tables (0.64).
Answer for c)
The mean of the modified data set = 30+30+32+34 / 4 = 31.5
Answer for d)
Estimated standard deviation of the modified data set:
Standard deviation, = [ (x-mean)2 / N ]
Accordingly
= (30 – 31.5)2 + (30 – 31.5)2 + (32 – 31.5)2 + (34 – 31.5)2 / 4
= (-1.5)2 + (-1.5)2 + (0.5)2 + (2.5)2 / 4
= 2.25 + 2.25 + 0.25 + 6.25 / 4
= 2.75
Standard deviation, = Square root of 2.75 = 1.65
Estimated standard deviation of the modified data set = 1.65
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