The results from two different lab sections for the first part of the second exp
ID: 3292953 • Letter: T
Question
The results from two different lab sections for the first part of the second experiment (NaOH concentration standardization using KHP) are given in the table below. {For this question, treat each student’s result as a single measurement; since we don’t actually know how many trials the student conducted.}
A) Identify the one most likely outlier in each data set, and test each to ascertain whether it could be rejected on statistical grounds. If you decide to reject data, make sure that you state the confidence level (and critical value) that you used and then be sure to recalculate the mean.
B) State the “answer” for each lab in the most complete, scientifically correct way that you have been taught (i.e., write the confidence intervals). {Note: the standard deviations for the two sets drop to 0.0113 and 0.0400 if you take out the top number in each set.}
C) If you were in a rush on a test and didn’t want to go to all the trouble of calculating a pooled standard deviation for the two sets of data above, you could estimate spooled from the individual data set’s standard deviations using any one of a number of methods we discussed in class. Quote a value for this estimated value of spooled and justify why you chose it a sentence or less.
D) A potential problem with the estimated value above OR with calculating a proper value of spooled is that it is only reasonable to construct a pooled standard deviation if the data sets have statistically indistinguishable standard deviations. Perform a test to see if this is the case and explain what the result means in light of the test performed below in part E.
E) Given that we used to have two carboys (before we got the really big ones) in the lab and intentionally used different concentrations of acid and base for the different lab sections, make a statement about whether these two sets of results are likely to be from the same carboy or not.
{Be sure to state your problem in proper statistical terms, perform the appropriate statistical test, make a statistical conclusion, and finally give the laymen’s version of the answer to the question above.}
Lab 1
Lab 2
0.9022
0.712
0.941
0.826
0.943
0.932
0.9439
0.935
0.9453
0.946
0.9501
0.9462
0.9502
0.9468
0.955
0.948
0.9631
0.9522
0.976
0.9575
0.962
0.973
0.983
0.9903
Mean
0.9470
0.9293
SD
0.0190
0.0734
A) Identify the one most likely outlier in each data set, and test each to ascertain whether it could be rejected on statistical grounds. If you decide to reject data, make sure that you state the confidence level (and critical value) that you used and then be sure to recalculate the mean.
B) State the “answer” for each lab in the most complete, scientifically correct way that you have been taught (i.e., write the confidence intervals). {Note: the standard deviations for the two sets drop to 0.0113 and 0.0400 if you take out the top number in each set.}
C) If you were in a rush on a test and didn’t want to go to all the trouble of calculating a pooled standard deviation for the two sets of data above, you could estimate spooled from the individual data set’s standard deviations using any one of a number of methods we discussed in class. Quote a value for this estimated value of spooled and justify why you chose it a sentence or less.
D) A potential problem with the estimated value above OR with calculating a proper value of spooled is that it is only reasonable to construct a pooled standard deviation if the data sets have statistically indistinguishable standard deviations. Perform a test to see if this is the case and explain what the result means in light of the test performed below in part E.
E) Given that we used to have two carboys (before we got the really big ones) in the lab and intentionally used different concentrations of acid and base for the different lab sections, make a statement about whether these two sets of results are likely to be from the same carboy or not.
{Be sure to state your problem in proper statistical terms, perform the appropriate statistical test, make a statistical conclusion, and finally give the laymen’s version of the answer to the question above.}
Explanation / Answer
Solution ± A µ2 [1,n]
Let X = Lab 1 result and Y = Lab 2 result. We assume X ~ N(µ1, 12) and Y ~ N(µ2, 22)
Back-up Theory
Part (a) and (b)
Outliers
99% confidence interval for individual observations is: Mean ± 2.576s, where 2.576 is the upper 0.5% point of N(0, 1) and s = sample standard deviation.
Using the above formula, 99% CI for Lab 1 results: 0.9470 ± (2.576 x 0.0190)
= 0.9470 ± 0.0489= (0.8981, 0.9959)
Since none of the results falls outside this interval, Lab 1 results do not have any outlier.
Similarly, 99% CI for Lab 1 results: 0.9293 ± (2.576 x 0.0734)
= 0.9470 ± 0.1891= (0.7579, 1.1361)
Since the result 0.712 < lower bound above, it is an outlier.
Discarding this result, the revised mean = {(0.9470 x 10) – 0.712}/9 = 0.9731.ANSWER
Part (c)
To test H0 : µ1 = µ2 Vs H1 : µ1 µ2
Pooled variance ={ (n1 - 1)s12 + (n2 - 1)s22 }/( n1 + n2 - 2)
= {(9 x 0.01902) + (13 x 0.07342)}/(10 + 14 - 2) = 0.00333124
So, pooled standard deviation, s = 0.00333124 = 0.0577.
Test Statistic: t = (Xbar - Ybar)/[s{(1/n1) + (1/n2)}]
= (0.9470 - 0.9731)/ [0.0577{(1/10) + (1/14)}] = - 0.0261/0.0239 = 1.092.
p-value = P(t13 > | 1. 092 |) = 0.2947 [under H0 , t ~ t n1 + n2 - 2]
Since p-value is much higher than even 0.1, H0 is accepted.
=> the two means are equal.
Pooled standard deviation is preferred to standard deviation of pooled set of observations on two counts:
1. Standard deviation of pooled set of observations has an inflationary tendency when the respective means are not very close to each other.
2. If 1 = 2 = , say, then both s1 and s2 are unbiased estimators of and average of two unbiased estimators always leads to a better estimate.
DONE
Part (d)
To test 1 = 2, test statistic is: F = s22/s12 = 0.0734/0.0190 = 3.863 [note that F is the ratio of larger variance to smaller variance]
p-value = P(F9. 13 > 3.863) = 0.0139 [under H0 , F ~ F n1 – 1, n2 - 1]
Since p-value is larger than 0.01, H0 is accepted.
=> the two standard deviations are equal.
DONE
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