Research Methodology - Dr. Krishan K. Pandey
Research Methodology - Dr. Krishan K. Pandey Research Methodology - Dr. Krishan K. Pandey
1 2 3 4 5 z = p q 0 0 p$ − p$ F HG 1 2 1 1 + n n 1 2 I KJ and q = 1 − p in which case 0 0 we calculate test statistic variance Population(s) χ 2 -test and the test F-test and the test statistic 2 eσ pj normal, observations are independent statistic 2 2 σ ∑ X X n s i − − 1 d 1 1i / 1 F = = 2 2 σs2 ∑d X 2i − X2i / n − 1 2 2 σ s χ = 2 σ bn − 1g 2 2 where σs1 is treated > σ s2 In the table the various symbols stand as under: p with d.f. = (n – 1) with d.f. = v 1 = (n 1 –1) for greater variance and d.f. = v 2 = (n 2 – 1) for smaller variance X = mean of the sample, X 1 = mean of sample one, X 2 = mean of sample two, n = No. of items in a sample, n 1 = No. of items in sample one, n 2 = No. of items in sample two, μ H0 = Hypothesised mean for population, σ p = standard deviation of population, σ s = standard deviation of sample, p = population proportion, q = 1 − p , p$ = sample proportion, q$ = 1 − p$ . Testing of Hypotheses I 201
202 Research Methodology and σ s = ∑ − 2 dXiXi bn − 1g 5. Population may not be normal but sample size is large, variance of the population may be known or unknown, and H a may be one-sided or two-sided: In such a situation we use z-test and work out the test statistic z as under: X −μH z = σ / n p (This applies in case of infinite population when variance of the population is known but when variance is not known, we use σ s in place of σ p in this formula.) OR X − μH0 z = eσp/ nj × bN − ng/ bN − 1g (This applies in case of finite population when variance of the population is known but when variance is not known, we use σ s in place of σ p in this formula.) Illustration 2 A sample of 400 male students is found to have a mean height 67.47 inches. Can it be reasonably regarded as a sample from a large population with mean height 67.39 inches and standard deviation 1.30 inches? Test at 5% level of significance. Solution: Taking the null hypothesis that the mean height of the population is equal to 67.39 inches, we can write: H0 : μ H = 67. 39" 0 Ha: μH ≠ 67. 39" 0 and the given information as X = 67. 47", σ p = 130 . ", n = 400. Assuming the population to be normal, we can work out the test statistic z as under: X − μH67. 47 − 67. 39 008 0 . z = = = = 1231 . σ / n 130 . / 400 0. 065 p As H is two-sided in the given question, we shall be applying a two-tailed test for determining the a rejection regions at 5% level of significance which comes to as under, using normal curve area table: R : | z | > 1.96 The observed value of z is 1.231 which is in the acceptance region since R : | z | > 1.96 and thus H is accepted. We may conclude that the given sample (with mean height = 67.47") can be regarded 0 0
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1 2 3 4 5<br />
z =<br />
p q<br />
0 0<br />
p$ − p$<br />
F<br />
HG<br />
1 2<br />
1 1<br />
+<br />
n n<br />
1 2<br />
I<br />
KJ<br />
and q = 1 − p in which case<br />
0 0<br />
we calculate test statistic<br />
variance Population(s) χ 2 -test and the test F-test and the test statistic<br />
2 eσ pj normal, observations<br />
are independent<br />
statistic<br />
2<br />
2<br />
σ ∑ X X n<br />
s<br />
i − −<br />
1 d 1 1i<br />
/ 1<br />
F = =<br />
2<br />
2<br />
σs2<br />
∑d X 2i − X2i / n − 1<br />
2<br />
2 σ s<br />
χ =<br />
2<br />
σ<br />
bn − 1g<br />
2 2<br />
where σs1 is treated > σ s2<br />
In the table the various symbols stand as under:<br />
p<br />
with d.f. = (n – 1) with d.f. = v 1 = (n 1 –1) for<br />
greater variance and<br />
d.f. = v 2 = (n 2 – 1) for smaller variance<br />
X = mean of the sample, X 1 = mean of sample one, X 2 = mean of sample two, n = No. of items in a sample, n 1 = No. of items in sample one,<br />
n 2 = No. of items in sample two, μ H0 = Hypothesised mean for population, σ p = standard deviation of population, σ s = standard deviation of<br />
sample, p = population proportion, q = 1 − p , p$<br />
= sample proportion, q$ = 1 − p$<br />
.<br />
Testing of Hypotheses I 201