To test the effect of a fertilizer on rice production, 24 equal plots of a certain land are selected. Half of them were treated with fertilizer leaving the rest untreated. Other conditions were the same. The mean production of rice on untreated plots was 4.8 quintals with standard deviation of 0.4 quintal, while the mean yield on the treated plots was 5.1 quintals with a standard deviation of 0.36 quintal. Can we say that there is significant improvement in the production of rice due to use of fertilizer at 0.05 level of significance ?
my chi square test value = 5.67 , but teh chi sqaure critical at α = 0.05 and v = 24 is 36.415 , i do n't have enough evidenvce to reject H0 ,
H0 = 7.2
H1= <7.2 ,
But the ans provided is REJECT H0 , why ?
testing of population varience?
2017-11-29 7:31 pm
回答 (1)
2017-11-29 8:37 pm
This is a problem of test of difference between two means (not variances). We use the t-test (not the chi-squared
test)
I also don't see where you obtained 7.2 from.
First, test for the equality of population variances in the two sample to choose the appropriate t-test.
H0: σ (1)^2 = σ (2)^2
Ha: σ (1)^2 ≠ σ (2)^2
Larger variance = 0.16
Smaller variance = 0.1296
F = 1.23457
Degrees of freedom 11 and 11
Critical F(11,11,0.05) = 2.8536
Calculated F is smaller than critical F so don't reject H0. There is no evidence of difference between the two
variances. We choose to use the pooled-variance t-test.
H0: μ (1) = μ (2)
HA: μ (1) < μ (2)
Sample 1 size 12
Sample 2 size 12
Sample 1 mean 4.8
Sample 2 mean 5.1
Sample 1 S.D. 0.4
Sample 2 S.D. 0.36
Pooled S.D = [(n1-1)s1^2+(n2-1)s2^2]/(n1+n2-2)
Pooled variance s^2 = [(11)(0.16)+(11)(0.1296))] / (22) = 0.1448
Standard error of difference in means = sqrt(1/n1+1/n2) times sqrt(s^2)
Standard error of difference in means = (0.408248)(0.380526) = 0.155349 (denominator of t)
t = (mean1-mean2) / SE = (4.8 - 5.1) /0.155349
t = -1.931135
Degrees of freedom = 22
The critical |t| from the t-table with 22 degrees of freedom with level of significance 0.05 is 1.717
calculated |t| = 1.931
Calculated |t| exceeds critical t so reject H0.
we can say that there is significant improvement in the production of rice due
to use of fertilizer at 0.05 level of significance
test)
I also don't see where you obtained 7.2 from.
First, test for the equality of population variances in the two sample to choose the appropriate t-test.
H0: σ (1)^2 = σ (2)^2
Ha: σ (1)^2 ≠ σ (2)^2
Larger variance = 0.16
Smaller variance = 0.1296
F = 1.23457
Degrees of freedom 11 and 11
Critical F(11,11,0.05) = 2.8536
Calculated F is smaller than critical F so don't reject H0. There is no evidence of difference between the two
variances. We choose to use the pooled-variance t-test.
H0: μ (1) = μ (2)
HA: μ (1) < μ (2)
Sample 1 size 12
Sample 2 size 12
Sample 1 mean 4.8
Sample 2 mean 5.1
Sample 1 S.D. 0.4
Sample 2 S.D. 0.36
Pooled S.D = [(n1-1)s1^2+(n2-1)s2^2]/(n1+n2-2)
Pooled variance s^2 = [(11)(0.16)+(11)(0.1296))] / (22) = 0.1448
Standard error of difference in means = sqrt(1/n1+1/n2) times sqrt(s^2)
Standard error of difference in means = (0.408248)(0.380526) = 0.155349 (denominator of t)
t = (mean1-mean2) / SE = (4.8 - 5.1) /0.155349
t = -1.931135
Degrees of freedom = 22
The critical |t| from the t-table with 22 degrees of freedom with level of significance 0.05 is 1.717
calculated |t| = 1.931
Calculated |t| exceeds critical t so reject H0.
we can say that there is significant improvement in the production of rice due
to use of fertilizer at 0.05 level of significance
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