Research Methodology - Dr. Krishan K. Pandey
Research Methodology - Dr. Krishan K. Pandey Research Methodology - Dr. Krishan K. Pandey
Testing of Hypotheses I 205 S. No. X i Xi X − d i Xi − X d i 2 4 568 – 4 16 5 572 0 0 6 578 6 36 7 570 – 2 4 8 572 0 0 9 596 24 576 10 544 – 28 784 d i i 2 n = 10 ∑ Xi = 5720 ∑ X − X = 1456 ∴ X and σ s = ∑ − Hence, X = n ∑ dXiXi 2 n − 1 i 5720 = = 10 = 572 kg. 1456 = 10 − 1 572 − 578 t = =−1488 . 12. 72/ 10 12. 72 kg. Degree of freedom = (n – 1) = (10 – 1) = 9 As H a is two-sided, we shall determine the rejection region applying two-tailed test at 5 per cent level of significance, and it comes to as under, using table of t-distribution * for 9 d.f.: R : | t | > 2.262 As the observed value of t (i.e., – 1.488) is in the acceptance region, we accept H 0 at 5 per cent level and conclude that the mean breaking strength of copper wires lot may be taken as 578 kg. weight. The same inference can be drawn using Sandler’s A-statistic as shown below: Table 9.3: Computations for A-Statistic S. No. Xi Hypothesised mean mH = 578 kg. 0 Di = Xi − μH0 d i D i 2 1 578 578 0 0 2 572 578 –6 36 3 570 578 –8 64 4 568 578 –10 100 contd. ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ * Table No. 2 given in appendix at the end of the book.
206 Research Methodology S. No. Xi Hypothesised mean mH = 578 kg. 0 Di = Xi − μH0 d i D i 2 5 572 578 –6 36 6 578 578 0 0 7 570 578 –8 64 8 572 578 –6 36 9 596 578 18 324 10 544 578 –34 1156 2 n = 10 ∑ Di = −60 ∑ Di = 1816 b g b g 2 2 2 i i ∴ A =∑D / ∑ D = 1816/ − 60 = 05044 . Null hypothesis H : μ H = 578 kg. 0 0 Alternate hypothesis Ha: μH ≠ 578 kg. 0 As H a is two-sided, the critical value of A-statistic from the A-statistic table (Table No. 10 given in appendix at the end of the book) for (n – 1) i.e., 10 – 1 = 9 d.f. at 5% level is 0.276. Computed value of A (0.5044), being greater than 0.276 shows that A-statistic is insignificant in the given case and accordingly we accept H 0 and conclude that the mean breaking strength of copper wire’ lot maybe taken as578 kg. weight. Thus, the inference on the basis of t-statistic stands verified by A-statistic. Illustration 6 Raju Restaurant near the railway station at Falna has been having average sales of 500 tea cups per day. Because of the development of bus stand nearby, it expects to increase its sales. During the first 12 days after the start of the bus stand, the daily sales were as under: 550, 570, 490, 615, 505, 580, 570, 460, 600, 580, 530, 526 On the basis of this sample information, can one conclude that Raju Restaurant’s sales have increased? Use 5 per cent level of significance. Solution: Taking the null hypothesis that sales average 500 tea cups per day and they have not increased unless proved, we can write: H : μ = 500 cups per day 0 H a : μ > 500 (as we want to conclude that sales have increased). As the sample size is small and the population standard deviation is not known, we shall use t-test assuming normal population and shall work out the test statistic t as: t X −μ = σ / n (To find X and σ s we make the following computations:) s
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206 <strong>Research</strong> <strong>Methodology</strong><br />
S. No. Xi Hypothesised mean<br />
mH = 578 kg.<br />
0<br />
Di = Xi<br />
− μH0 d i D i 2<br />
5 572 578 –6 36<br />
6 578 578 0 0<br />
7 570 578 –8 64<br />
8 572 578 –6 36<br />
9 596 578 18 324<br />
10 544 578 –34 1156<br />
2<br />
n = 10 ∑ Di = −60<br />
∑ Di = 1816<br />
b g b g<br />
2 2 2<br />
i i<br />
∴ A =∑D / ∑ D = 1816/ − 60 = 05044 .<br />
Null hypothesis H : μ H = 578 kg.<br />
0 0<br />
Alternate hypothesis Ha: μH ≠ 578 kg.<br />
0<br />
As H a is two-sided, the critical value of A-statistic from the A-statistic table (Table No. 10 given<br />
in appendix at the end of the book) for (n – 1) i.e., 10 – 1 = 9 d.f. at 5% level is 0.276. Computed<br />
value of A (0.5044), being greater than 0.276 shows that A-statistic is insignificant in the given case<br />
and accordingly we accept H 0 and conclude that the mean breaking strength of copper wire’ lot<br />
maybe taken as578 kg. weight. Thus, the inference on the basis of t-statistic stands verified by<br />
A-statistic.<br />
Illustration 6<br />
Raju Restaurant near the railway station at Falna has been having average sales of 500 tea cups per<br />
day. Because of the development of bus stand nearby, it expects to increase its sales. During the first<br />
12 days after the start of the bus stand, the daily sales were as under:<br />
550, 570, 490, 615, 505, 580, 570, 460, 600, 580, 530, 526<br />
On the basis of this sample information, can one conclude that Raju Restaurant’s sales have increased?<br />
Use 5 per cent level of significance.<br />
Solution: Taking the null hypothesis that sales average 500 tea cups per day and they have not<br />
increased unless proved, we can write:<br />
H : μ = 500 cups per day<br />
0<br />
H a : μ > 500 (as we want to conclude that sales have increased).<br />
As the sample size is small and the population standard deviation is not known, we shall use t-test<br />
assuming normal population and shall work out the test statistic t as:<br />
t<br />
X −μ<br />
=<br />
σ / n<br />
(To find X and σ s we make the following computations:)<br />
s