Chemometrics. Matti Hotokka Physical chemistry Åbo Akademi University

Transcription

1 Chemometrics Matti Hotokka Physical chemistry Åbo Akademi University

2 Hypothesis testing Inference method Confidence levels Descriptive statistics Hypotesis testing Predictive statistics

3 Hypothesis testing The steps involved Formulate a null hypotesis This is what you want to claim E.g., the sample is within tolerances Formulate an alternative hypotesis This is a complement to null hypotesis E.g., the sample is not within tolerances Calculate a characteristic number Compare with tabulated values Accept or reject the null hypotesis

4 Hypothesis testing Huge number of tests exist Tests for mean Tests for distribution Tests for spread Tests for outliers Etc.

5 Hypothesis testing Test for the mean Double-sided t-test, x = ì P(X) Acceptable No-no No-no x

6 Hypothesis testing Mean at nominal value (double-sided) The ibuprofen concentration must be 400 mg per pill. Therefore ì = 400 mg. Take 5 pills and measure the ibuprofen content. The results are 396, 388, 398, 382, 373 mg. Mean x = 387 mg, s = 10.3 mg. Calculate the critical number, t = 2.82 Degrees of freedom = n-1 = 4 Choose risk level: 5 % (95 % confidence) Read the table for Student s t-test at risk level because the risk 2.5 % at the low end and 2.5 % at the high end gives total risk of 5 %. The value in the table, 2.776, is smaller than the calculated one. Reject the null hypotesis. Accept the alternative hypotesis. We cannot guarantee at 95 % confidence level that the pills have the prescribed amount of ibuprofen.

7 Student s distribution Reminder D.f. Risk N = number of samples D.f. = degrees of freedom = N - 1 This table is one-sided. Therefore the total risk at level is 2.5 % % and confidence probability is 95 %.

8 Hypothesis testing Test for the mean One-sided t-test, x = ì P(X) Acceptable ì No-no x

9 Hypothesis testing Mean below a nominal value (one-sided) The EU regulatory limit for nitrate in drinking water is 50 mg/l. Determinations from 4 parallel samples gave the results 51.0, 51.3, 51.6, 50.9 mg/l. Is this just random variation or is the observed level systematically above the prescribed limit? Mean 51.2 and st.dev mg/l. Null hypotesis: the level is not exceeded, x ì, alternative hypotesis: it is too high. Calculate t = Choose risk level: 5 %. D.f. = 4-1 = 3. The tabulated value of t, 2.353, is smaller than the calculated one. The null hypotesis must be rejected. The concentration is too high.

10 Hypothesis testing Compare two means Compare two sets of parallel measurements from different samples. Do the two samples differ significantly? A two-sided test.

11 Hypothesis testing Do two production batches differ? Quality control tests the day and night shifts at a refinery. The octan numbers of parallel measurements are (1: day) 94.92, 95.07, 94.96, 95.02, 94.99, 94.93; (2: nite) 95.03, 95.08, 94.98, 95.03, 95.01, Means: (1) 94.98; (2) St.dev.: (1) 0.057; (2) Weighted st.dev. = Student s t = d.f. = 10 Choose risk level 2.5 %, read column : t = Comparison: No, we cannot say that the two results differ. Therefore only random variations are observed.

12 Hypothesis testing Dixon s Q test for outliers Can be applied also for very few observations. Arrange your n observations in ascending order. Calculate the numbers Q 1 and Q n. Null hypotesis: not an outlier. Accepted if calculated Q less than tabulated.

13 Hypothesis testing Dixon s Q test for outliers Critical values of Q test at the 1 % risk level. Number of observations = n. n Q n Q

14 Hypothesis testing Dixon s Q test for outliers Persons of the following ages participate in a bus tour to see a theater performance in Helsinki: 6, 7, 5, 6, 7, 6, 103, 8, 7, 5. Order them: 5, 5, 6, 6, 6, 7, 7, 7, 8, 103. Q 1 = 0, 5 is not an outlier; Q n = 0.969, 103 certainly is an outlier.

15 Hypothesis testing Grubb s test for outliers Observation x* is not an outlier in a series if

16 Hypothesis testing Grubb s test for outliers Critical values for Grubb s outlier test at 95 % and 99 % levels. Number of observations = n. n T(95%) T(99%) n T(95%) T(99%)

17 Hypothesis testing Outliers in linear regression In order to find whether or not observation k (value y k ) is an outlier 1) Calculate a new regression with observation k removed. 2) Calculate e k = y k obs - y k calc. 3) Reject if distance exceeds a set limit; the limit is often two or three standard deviations

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19 ANOVA Analysis of variance Used to test interdependences between batches. Used as an analysis tool for designed experiments. Requires several parallel measurements (replicates) in each batch (or experiment).

20 ANOVA One-way analysis Assume that samples are taken at four different times from waste water of a factory to study the potassium concentration (mg/l). Each sample is analysed by a different crue. Three parallel measurements are made to determine the concentration of each sample. Replicate Batch Mean Do the true concentrations ì 1, ì 2, ì 3 and ì 4 differ?

21 ANOVA Variation between samples Replicate Batch Mean y total = SSQ fact = 0.494

22 ANOVA Variation between samples Replicate Batch Mean y total = Number of batches SSQ fact = Number of replicates

23 ANOVA Variation between samples Yeah, well the variance is So what? The variance of the means must be related to the general fluctuations in the whole set of measurements. Calculate the spread in batch 1, 2 etc and combine. Then there must be a test to see whether or not a critical value at a given risk level is exceeded.

24 ANOVA Variation between samples Replicate Batch Mean y total = SSQ fact = SSQ R = Ó(y ij - y j ) 2

25 ANOVA Variation between samples Replicate Batch Mean y total = SSQ fact = SSQ R = Ó(y ij - y j ) 2

26 ANOVA Variation between samples Calculate the variance of the means (á is the number of batches) Calculate the pooled variance of the whole set Form the test quantity

27 ANOVA Variation between samples Compare the calculated F value with the tabulated values to find out how probable it is that the two variances differ so much. Degrees of freedom: Numerator: á - 1 = 4-1 = 3 Denominator: á(n - 1) = 4(3-1) = 8 The F value is beyond the critical value F 0.01 = 7.59 so there is a less than 1 % chance that the concentrations are the same. Blow the whistle.

28 F distribution Df for denom Df for num F F F F F F F F F F F F F F F Critical value E.g., F 0.05 means that there is a 5 % probability variances differ this much. Choose the correct df s and read down the column until you reach a probability level that matches your critical value.

29 ANOVA One-way analysis Another view on F: Of all possible sources of difference, only the difference of the batches is considered, hence one-way. The model is the average of all batches, The spread explained by the model is the fluctuation of individual batch averages from the overall average. This is s X2. The unexplained part is the fluctuation of the individual observations from the corresponding batch averages. This is measured by s p2.

30 ANOVA Two-way analysis There is another factor influencing the fluctuations besides the difference in batches, namely the difference in determinations in individual replications. In a two-way analysis both are considered.

31 ANOVA Two-way analysis Replicate Batch Replicate means Mean The one-way model predicts the overall mean, The two-way analysis predicts for observation batch=2, replicate=1 the overall mean plus a correction for batch mean ( =0.28) plus a correction for the replica mean ( =-0.09). Thus the model predicts the value = The residual fluctuation is now 0.01.

32 ANOVA Two-way analysis Replicate Batch X (X j -X ) X i (X i -X ) #Replicates SS A = 3* = Differences between batches, Df = 4-1 = 3 SS B = 4* = Differences between replicates, Df = 3-1 = 2 #Batches

33 ANOVA Two-way analysis Residuals Replicate Batch Predicted value Residual SS R = , Df = 3*2 = 6

34 ANOVA Two-way analysis For the batches For the replications For the residuals For the batches For the replications

35 ANOVA Two-way analysis The critical value in the table shows that the probability that the batches are similar is less much than 1 %. The probability that the replicates are similar is roughly 25 %.

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