# 99 per cent confidence limits of the mean height of

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99 per cent confidence limits of the mean height of the population, x Z SE 63 2.58 (0.3) 0.774 62.20 and 63.80 Illustration - 3 A sample of 900 items is taken from a normal population whose mean is 6 and variance is 6. If the sample mean is 6.60 can the sample be regarded as a truly random sample. Give necessary justification for your conclusions.

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90 Solution Let us take the hypothesis that there is no difference between the sample mean and the population mean. SE x = S/ n 900 45 . 2 x SE 45 . 2 6 Variance S = 0.0817 Difference between the sample mean and population mean 6.60 6.00 = 0.60 0817 . 0 0.60 SE Difference population of mean (Expected Z Since the difference is more than 1.96 SE at 0.05 level, it could not have arisen due to variations of sampling. Hence, the sample cannot be regarded s truly random sample. Illustration - 4 If it costs a rupee to draw one number of a sample, how much would it cost in sampling from a universe with mean 100 and standard deviation 9 to take sufficient number as to ensure that the mean of a sample would be within 0.015 per cent of the value at 0.05 level. Find the extra cost necessary to double this precision. Solution We are given, X = 100 and S.D = 9 Difference between sample mean and population mean = 0.015 (given) For 95 per cent confidences, difference between sample mean and population means should be equal to 1.96 SE 0.015 = 1.96 SE n S SE 015 . 0 n S 1.96 Therefore, n 015 . 0 9 96 . 1 015 . 0 n 9 1.96
91 015 . 0 9 1.96 n sides) both squaring (by (1176) n n = (1176) 2 = 13,82,976 The difference between sample mean and population mean by making the precision double, it should be = 0.015 /2.0 = 0.0075. So, 0075 . 0 n 9 1.96 sides) both squaring (by 0.0075 9 1.96 n n = 55,31,904 Therefore, Extra cost = 55,31,904 - 13,82,976 = Rs.41,48,928 Hence to double the precision, the extra cost would be Rs.41,48,928 Illustration - 5 The average number of defective articles in a factory is claimed to be less than for all the factories whose average is 30.5. A random sample of 100 defective articles showed the following distribution. Class limits Number 16-20 21-25 26-30 31-35 36-40 12 22 20 30 16 Calculation the mean and standard deviation of the sample and use it to test the claim that the average is less than the figure for all the factories at 0.05 level of significance. Solution The sample mean and population mean do not differ i.e. Ho : x = Class limits Mid value Dx = x-28/5 f fdx f.dx 2 16-20 21-25 26-30 31-35 36-40 18 23 28 33 38 -2 -1 0 +1 +2 12 22 20 30 16 -24 -22 0 +30 +32 48 22 0 30 64

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92 n c fdx A x 100 5 16 28 = 28 + 0.8 = 28.8 c n fdx n x fd S 2 2   5 100 16 100 164 2 5 6144 . 1 5 0256 . 0 64 . 1 = 6.35 Test Statistic n S/ - x x 6.35 17.00 - 6.35 100 1.7 - 100 6.35/ 30.5 - 28.8 = -2.68 Since Z is more than 1.96, it is significance at 0.05 the level of significance. Hence we reject the null hypothesis and conclude that the sample mean and population mean differ significantly. IN other words, the manufacturer’s claim that the average number of defectives in his product is less than the average figure for all the factories is valid.
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