![Test of hypothesis on the difference
Test of hypothesis on the difference
between 2 proportions (Ho: P
between 2 proportions (Ho: P1
1 = P
= P2
2)
)
n = [Z(1 – α) √{2P (1 – P)} + Z(1 – β) √{P1(1 – P1)} + P2 (1 – P2)]2
(P2 – P2
2)2
Where P = (P1 + P2 ) / 2
P1= Proportion of first group
P2= Proportion of second group
Z(1 – α) = z deviate corresponding to the α
error rate
Z(1 – β) = z deviate corresponding to the β
error rate](https://image.slidesharecdn.com/determinationofsamplesize-250513011104-a6703864/85/what-are-the-Determination-of-sample-size-ppt-14-320.jpg)
![P = (P1 + P2 )/ 2 = 0.70
P1= Proportion of first district = 0.80
P2= Proportion of second district = 0.20
Z(1 – α) = z deviate corresponding to the α
error rate = 1.282
Z(1 – β) = z deviate corresponding to the β
error rate = 0.842
n = [1.282 √{2(0.70) (0.30)} + 0.842 √{(0.80) (0.20)} + (0.60) (0.40)]2
(0.80 – 0.60)2
= 46.47 = 47 children is sample size](https://image.slidesharecdn.com/determinationofsamplesize-250513011104-a6703864/85/what-are-the-Determination-of-sample-size-ppt-16-320.jpg)
what are the Determination of sample size.ppt
- 1. Determination of Sample Determination of Sample Size Size
- 2. Determination of Sample Size Determination of Sample Size Most of the researcher are not sure of how many sample they need, to collect the information. Most the text books advocate to use as large sample as possible. However that is impossible. On the other hand there is a approach including of one tenth of total population. The other methods for sample estimation is use of the formulas which depends on many factors. The first and foremost is the type of study to be undertaken, whether observational or experimental.
- 3. For cross sectional studies For cross sectional studies For the population mean, μ of sample drawn by simple random sampling method n = Zα 2 s2 d2 Where n = required sample size Zα = z deviate corresponding to the reliability level s2 = variance ( s – standard deviation ) d = maximum tolerable error.
- 4. A nutritionist wishes to conduct a survey among teenage girls to determine their average daily protein intake. If she would like her estimated to be within 5 units of the true value with a reliability level of 95%, how many girls should be taken into the study if the standard deviation of protein in 20?
- 5. Calculation of Sample Size Calculation of Sample Size Z = 1.96 for 95% reliability D = 5 units = tolerable error S = 20 units = standard deviation ( Variance = 202 ) Therefore required sample size n = {(1.96)2 X 202 } / 52 = 61.5 = 62
- 6. For the population proportion, P –simple random sample n = Zα 2 PQ d2 Where n = required sample size Zα = z deviate corresponding to desired reliability level P = estimated proportion in the population Q = 100 – P ( if P is in % ) d = maximum tolerable error = 10% of P
- 7. • A survey to determine the prevalence rate of TB is to be undertaken. How many subjects should be included in the study in the prevalence in the past is 35% and the desired precision and reliability are 10% and 95% respectively?
- 8. Calculation of sample size Calculation of sample size Z = 1.96 for 95% reliability P = 35% =given proportion Q = 100 – 35 ( if P is in % ) = 65% d = maximum tolerable error = 10% of P = 3.5 Therefore n = {(1.96)2 X 35X65 } / (3.5)2 = 713.44 = 714
- 9. For experimental studies For experimental studies • Experimental design • Magnitude of α error • Magnitude of β error • Minimum difference which has to be detected • Difference being detected or difference on proportions
- 10. Test of hypothesis on the difference between means (Ho: μ 1 = μ 2) n = 2(Zα + Zβ)2 s2 d2
- 11. Where: n = sample size required / group Zα = z deviate corresponding to the α error rate Zβ = z deviate corresponding to the β error rate s2 = variance d = difference to be detected
- 12. We wish to know whether a change in the structure of an analgesic drug increases the duration of pain relief. We specify a probability of 5% of failing to detect an increase of 15 min. with an error rate of 2 ½%. The standard deviation. Of the duration of pain relief, know for the old treatment from previous experience, is 1 hr. and we assume that it is the same for the new treatment.
- 13. Sample size calculation Sample size calculation Zα = 1.96 for α = 5% Zβ = 1.645 for β = 2 ½%. s = 1 hour d = difference to be detected = 1/4 hour n = 2(1.96 + 1.645)2 12 (1/4)2 = 416 subjects/group
- 14. Test of hypothesis on the difference Test of hypothesis on the difference between 2 proportions (Ho: P between 2 proportions (Ho: P1 1 = P = P2 2) ) n = [Z(1 – α) √{2P (1 – P)} + Z(1 – β) √{P1(1 – P1)} + P2 (1 – P2)]2 (P2 – P2 2)2 Where P = (P1 + P2 ) / 2 P1= Proportion of first group P2= Proportion of second group Z(1 – α) = z deviate corresponding to the α error rate Z(1 – β) = z deviate corresponding to the β error rate
- 15. Suppose it has been estimated that the rate is 800 per 1000 school children in one district and 600 per 1000 in another district. How large a sample of children form each district is required to determine whether this difference is significant the difference if it is real ?
- 16. P = (P1 + P2 )/ 2 = 0.70 P1= Proportion of first district = 0.80 P2= Proportion of second district = 0.20 Z(1 – α) = z deviate corresponding to the α error rate = 1.282 Z(1 – β) = z deviate corresponding to the β error rate = 0.842 n = [1.282 √{2(0.70) (0.30)} + 0.842 √{(0.80) (0.20)} + (0.60) (0.40)]2 (0.80 – 0.60)2 = 46.47 = 47 children is sample size
- 17. Thank You Thank You Very Much ! Very Much !