1. BASIC STATISTICAL INFERENCE A. COMPARE BETWEEN TWO MEANS OF POPULATIONS B. COMPARE BETWEEN TWO VARIANCES OF POPULATIONS PARAMETERIC TESTS (QUANTITATIVE DATA) t-distribution z-distribution f-distribution (fisher’s distribution) &

2. BASIC STATISTICAL INFERENCE We shall consider here three forms for the alternative hypothesis:

3. Not significant Distribution showing 0.05 significant level in one-tailed test 0.05 significant level 0.95 One tailed test P < 0.05 P < 0.01 P < 0.001 P > 0.05 Insignificant difference Not significant Distribution showing 0.05 significant level in one-tailed test 0.05 significant level 0.95 ?

4. Distribution showing 0.05 significant level in two-tailed test 0.05 significant level 0.95 Two tailed test Not significant

5. Frequen cy 0.05 0.01 Calculated t 0.001 Tabulated t

6. Accept H 0 Reject H 0 P > 0.05

7. Calculated z Mean sample A given fixed value to be tested Population standard deviation Sample size (>30) HYPOTHESIS TESTS ON THE MEAN (LARGE SAMPLES >30)

8. Calculated z Mean sample A given fixed value to be tested Sample standard deviation Sample size (<30) HYPOTHESIS TESTS ON THE MEAN (SMALL SAMPLES <30)

9. o To decide if a sample mean is different from a hypothesized population mean. o You have calculated mean value and standard deviation for the group assuming you have measurement data. where the standard score (t) is:standard One sample t-distribution Degree of freedom (n-1)

10. t-distribution o The percentiles values of the t-distribution (t p ) are tabulated for a range of values of d.f. and several values of p are represented in a Table.

11. The mean concentration of cadmium in water sample was 4 ppm for sample size 7 and a standard deviation=0.9 ppm. The allowable limit for this metal is 2 ppm. Test whether or not the cadmium level in water sample at the allowable limit. Example Solution T cal (2.447) > t tab (2.447) Reject the null hypothesis T cal (2.447) > t tab (3.707) Reject the null hypothesis Decision: Thus the cadmium level in water is not at the allowable limit. One sample t-DISTRIBUTION

12. Example: In an New Zealand, Does the average mass of male turtles in location A was significantly higher than Location B? Location ALocation B n2526 3835 S43 d.f. = n 1 + n 2 - 2 = 25 + 26 - 2 = 49 Tabulated t at df 59 = 1.671 Thus, t observed (3.35) > t tabulated (1.67) at α= 0.05 The mass of male turtles in location A is significantly higher than those of location B (reject H 0 ) P<0.05 Two sample INDEPENDENT t-DISTRIBUTION

13. Control (X 1 ) Pb (X 2 )7963 8371 6846 5957 8153 7646 8057 7476 5852 4968 73 (X 1 ) 2 6241 6889 4524 3481 6561 5776 6400 5476 3364 2401 4624 (X 2 ) 2 3969 5041 2116 3249 2809 2116 3249 5776 2704 4624 5329 7756625583740982 t calculated (2.209) > t tabulated (2.086) at d.f. 20 d.f. = n 1 + n 2 - 2 = 11+11 -2= 20

14. d.f. = n - 1 Before (X1)After (X 2 ) 140 60 43 1520 30 30 61 51 61 30 D -14 -6 5 -3 -5 -4 -5 -3 D2D2 196 36 1 25 9 9 16 25 9 d.f. = 10 – 1= 9 t tabulated at d.f. 10 = 1.833 ? -39351 TESTING THE DIFFERENCE BETWEEN TWO MEANS OF DEPENDENT SAMPLES Two sample DEPENDENT t-DISTRIBUTION

15. 1)That is, you will test the null hypothesis H 0 : σ 1 2 = σ 2 2 against an appropriate alternate hypothesis H a : σ 1 2 ≠ σ 2 2. 2)You calculate the F-value as the ratio of the two variances:  where s 1 2 ≥ s 2 2, so that F ≥ 1.  The degrees of freedom for the numerator and denominator are n 1 -1 and n 2 -1, respectively.  Compare F calc. to a tabulated value F tab. to see if you should accept or reject the null hypothesis. Fisher’s F-distribution

16. Example: Assume we want to see if a Method 1 for measuring the arsenic concentration in soil is significantly more precise than Method 2. Each method was tested ten times, with yielding the following values: MethodsMean (ppm)S.D. (ppm) Method 16.70.8 Method 28.21.2 So we want to test the null hypothesis H 0 : σ 2 2 = σ 1 2 against the alternate hypothesis H A : σ 2 2 > σ 1 2 Solution: o The tabulated value for d.f.= 9 in each case, at 1-tailed, 95% confidence level is F 9,9 = 3.179. o In this case, F calc 0.05 d.f.= 10 – 1 = 9 o We use a 1-tailed test in this case because the only information we are interested in is whether Method 1 is more precise than Method 2