JYOTI NIVAS COLLEGE AUTONOMOUS
DEPARTMENT OF BIOTECHNOLOGY
BIOSTATISTICS HYPOTHESIS TESTING - 3
Student’s t- test(solved problems with practice questions)
TABLE OF CONTENTS
Contents
Student’s t – test ______1
HYPOTHESIS TESTING
Student’s t- test
Some examples for unpaired ‘t – test’/ test for difference between two means (Fischer’s t- test) [ for two independent smaples]
This test is applied to unpaired data of independent observations made on individuals of two different groups or samples drawn from two populations.
4. Applications of fertilizers were tested for the yield of rice grown in 10 plots. Another seed of 10 plots of similar size and condition were taken as control. Test the effect of fertilizer.
Plot number 1 2 3 4 5 6 7 8 9 10 Fertilizer applied 16 14 18 15 13 17 16 15 14 13 Fertilizer not applied 10 12 11 9 13 13 12 14 13 11
Solution:
It is the case of small sample; hence t-test would apply.
Let the null hypothesis (H0): there is no significant effect of fertilizers on yield of rice grown
Alternative hypothesis (H1): there is significant effect of fertilizers on rice yield. Calculation: Let us fix the level of significance = 5% i.e. α = 0.05
ퟐ ퟐ Plot Fertilizer D1 = 푫ퟏ Fertilizer D2 = 푫ퟐ number applied X1 - 푿̅ퟏ not applied X2- 푿̅ퟐ (X1) (X2) 1 16 0.90 0.81 10 -1.8 3.24 2 14 -1.1 1.21 12 0.2 0.04 3 18 2.9 8.41 11 -0.8 0.64 4 15 -0.1 0.01 9 -2.8 7.84 5 13 -2.1 4.41 13 1.2 1.44 6 17 1.9 3.61 13 1.2 1.44 7 16 0.9 0.81 12 0.2 0.04 8 15 -0.1 0.01 14 2.2 4.84 9 14 -1.1 1.21 13 1.2 1.44 10 13 -2.1 4.41 11 -0.8 0.64 ퟐ ퟐ ΣX1= 151 Σ푫ퟏ = ퟐퟒ. ퟗퟎ ΣX2= 118 Σ푫ퟐ = ퟐퟏ. ퟔퟎ
Page 1 Shanmugam V. M. Dept. of Biotechnology, Jyoti Nivas College Autonomous HYPOTHESIS TESTING
ΣX1 ΣX2 푋̅1 = = 151/10 = 15.1 푋̅2 = = 151/10 = 15.1 푛1 푛2
푛1 = 10 푛2 = 10
2 2 Σ퐷1 = 24.90 Σ퐷2 = 21.60
∑(푋 −푋̅ )2 + ∑(푋 −푋̅ )2 SD (standard deviation) = √ 1 1 2 2 푛1+ 푛2 −2
2 2 퐷1 + 퐷2 = √ where, 퐷1 = 푋1 − 푋̅1; 퐷2= 푋2 − 푋̅2 푛1+ 푛2 −2
24.90 + 21.60 46.5 = √ = √ = = √2.584 = 1.604 10+ 10−2 18
SD = 1.604
ퟏ ퟏ Standard error (SE) of 푿̅ퟏ − 푿̅̅̅ퟐ̅ = 푺푫 √ + 풏ퟏ 풏ퟐ
ퟏ ퟏ SE = 1.604 √ + = 1.604 √ퟎ. ퟐ = 1.604 x 0.447 = 0.718 ퟏퟎ ퟏퟎ
Calculate the ‘t’ value i.e., the ratio between the observed difference and its standard error by substituting the above values in the formula:
|푋̅1− 푋̅̅̅2̅| |15.1 − 11.8| t calculated = = = 3.3/0.718 = 4.596 푆퐸 0.718
the level of significance is at 5% i.e., α = 0.05, critical value: the tabulated value of
t at α = 0.05 for df = (n1-1) + (n2-1) = n1 +n2 – 2 = 10+10-2 = 18 is 2.10
Decision: the calculated t value is more than the tabulated value i.e., tcal (4.596) >
ttab (2.10). so the null hypothesis is rejected i.e., there is significant difference between the control and fertilizer used for rice growth.
Assignment question 1:
Body lengths of fishes of a species was obtained from two ponds. They were measured as follows (in cms):
Pond A 20 24 20 28 22 20 24 32 24 26 Pond B 12 10 8 10 6 4 14 20 10 6
Calculate the mean difference in total body length between the two ponds of fish is significant or not (given tabulate t value at df 18 is 2.10)
Page 2 Shanmugam V. M. Dept. of Biotechnology, Jyoti Nivas College Autonomous HYPOTHESIS TESTING
PAIRED t - TEST
(for Dependent Samples where observation occur pairwise)
It is applied to paired data of dependent observations from one sample only. When each individual gives a pair observation.
1. To study the role of a factor or cause when the observations are made before and after its play. 2. To compare the effect of two drugs, given to the same person. 3. To compare the results of two different laboratory techniques 4. To compare observations made at two different sites in the same body.
WORKING PROCEDURE:
1. Fix the difference in each set of paired observations before and after (D = X1 –
X2). 2. Calculate the mean of the difference (퐷̅ ). 3. Work out the standard deviation (SD) of differences and then the standard error (SE) of the mean from the same .
2 (∑ 퐷) 푺푫 ∑ 퐷2− 푺푬 = where, SD = √ 푛 √(풏 ) 푛−1
4. Calculate the t value by substituting the above values in the formula |퐷̅ | |퐷̅ | t calculated = = 푆퐸 푺푫 √(풏 ) 5. Determine the degrees of freedom being one and the sample degree of freedom (df) should be n-1 6. Compare the calculate value with the table value at a particular degree of freedom (refer statistics table for t test).
Example:
5. Ten rats were fed with rice in first month and body weights of the rats were recorded. In the next month they were fed with grams and their weights were measured again. The respective weights of 10 rats in 2 months are as follows:
Wt. in 1st months 50 60 58 52 51 62 59 55 50 65 Wt. in 2nd months 56 58 68 61 56 59 64 60 50 62 Solution:
It is the case of small sample and same animal is given for different treatment; hence paired t-test would apply.
Page 3 Shanmugam V. M. Dept. of Biotechnology, Jyoti Nivas College Autonomous HYPOTHESIS TESTING
Let the null hypothesis (H0): Grains have no impact on rat’s nutrition
Alternative hypothesis (H1): Grains have impact on rat’s nutrition. Calculation: Let us fix the level of significance = 5% i.e. α = 0.05
st nd ퟐ Sl. No. Weights in 1 Weight in 2 D = X1- X2 푫 month (X1) month (X2) 1 50 56 50-56 = - 6 36 2 60 58 60-58 = 2 4 3 58 68 58-68 = -10 100 4 52 61 52-61 = -9 81 5 51 56 51-56 = -5 25 6 62 59 62 - 59 = 3 9 7 58 64 58 – 64 = -6 36 8 55 60 55 – 60 = -5 25 9 50 50 50-50 = 0 0 10 65 62 65 – 62 = 3 9 Σ푫 = −ퟑퟑ Σ푫ퟐ = ퟑퟐퟓ n = 10; Σ푫 = −ퟑퟑ ; Σ푫ퟐ = ퟑퟐퟓ
횺푫 −ퟑퟑ 퐷̅ = = = - 3.3 푛 10
2 (∑ 퐷) (−33)2 1089 ∑ 퐷2− 325− 325− 325− 108.9 SD = √ 푛 = SD = √ 10 = √ 10 = √ 푛−1 10−1 9 9
216.1 = √ 9 = √24.1 SD = ퟒ. ퟗ
푺푫 ퟒ.ퟗ ퟒ.ퟗ 푺푬 = = = = 1.55 √(풏 ) √ퟏퟎ ퟑ.ퟏퟔ
Calculation of t |퐷̅ | |퐷̅ | |−3.3 | t calculated = = = 2.13 = 푆퐸 푺푫 1.55 √(풏 ) 5% Level of significance = α is 0.05 Critical value: tabulated t value at α = 0.05 at degree of freedom (df) = n-1 = 10 – 1 = 9 is 2.26
Decision: the calculated t value i.e., t cal is 2.13 is less than tabulated t value (2.26). so null hypothesis is accepted that grams have no impact on nutrition in rats .
Page 4 Shanmugam V. M. Dept. of Biotechnology, Jyoti Nivas College Autonomous HYPOTHESIS TESTING
Assignment question 2:
2. Albino rats were administered with an Ayurvedic medicine at the rate of 10mg/10kg/day for 7 days. Initial and final body weights of the rats were recorded as shown in the table.
Rat number 1 2 3 4 5 6 7 8 9 10 Initial body weight 110 115 102 98 112 110 97 120 102 110 Final body weight 109 116 100 95 108 112 98 115 98 111
Determine whether the drug has any significant effect on the gain or loss of body weight of the rats? (Given tabulated t value at 0.05 df =10 -1 = 9 is 2.26)
Assignment question 3:
3. The fasting glucose level of seven persons were 4, 6, 8, 5, 9, 3 and 7mmol/L. after 75gram of oral glucose intake, the blood sugar level was 8, 10, 13, 9, 8, 12 & 14 mmol/L respectively. Use appropriate statistical test whether the oral glucose had significantly increased the blood sugar level or not? (Given tabulated t value at 0.05 df = 7 -1 = 6 is 2.45)
************
Page 5 Shanmugam V. M. Dept. of Biotechnology, Jyoti Nivas College Autonomous