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ssrs export to pdf barcode font Since n # 200 is large, we can assume that X is very nearly normal. in Software
Since n # 200 is large, we can assume that X is very nearly normal. Recognize QR Code JIS X 0510 In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. QR Code ISO/IEC18004 Creation In None Using Barcode generator for Software Control to generate, create QR Code JIS X 0510 image in Software applications. (a) The 95% confidence limits are # X or 0.824 1.96 s !n x # 1.96 s !n
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Painting Code39 In Java Using Barcode creator for Android Control to generate, create Code 3 of 9 image in Android applications. Printing UPC A In Java Using Barcode printer for Java Control to generate, create UPCA Supplement 2 image in Java applications. or 0.824 0.008 inches. Note that we have assumed the reported standard deviation to be the modified standard deviation ^. If the s standard deviation had been s, we would have used ^ !n>(n 1)s !200>199 s which can be taken as s s for all practical purposes. In general, for n 30, we may take s and ^ as practically equal. s Scan UPCA Supplement 2 In Visual Studio .NET Using Barcode scanner for Visual Studio .NET Control to read, scan read, scan image in .NET framework applications. Draw Bar Code In None Using Barcode encoder for Microsoft Excel Control to generate, create barcode image in Microsoft Excel applications. 6.7. Find (a) 98%, (b) 90%, (c) 99.73% confidence limits for the mean diameter of the ball bearings in Problem 6.6. Code 3/9 Scanner In None Using Barcode decoder for Software Control to read, scan read, scan image in Software applications. GS1128 Drawer In C# Using Barcode drawer for VS .NET Control to generate, create GS1128 image in VS .NET applications. CHAPTER 6 Estimation Theory
(a) Let zc be such that the area under the normal curve to the right of z zc is 1%. Then by symmetry the area to the left of z zc is also 1%, so that the shaded area is 98% of the total area (Fig. 61). Since the total area under the curve is one, the area from z 0 is z zc is 0.49; hence, zc 2.33. Therefore, 98% confidence limits are x # 2.33 s 2n 0.824 2.33 0.042 2200 0.824 0.0069 inch Fig. 61 Fig. 62 (b) We require zc such that the area from z Therefore, 90% confidence limits are x # 1.645 s 2n
0 to z
zc is 0.45; then zc 0.042 2200
1.645 (Fig. 62). 0.0049 inch
(c) The 99.73% confidence limits are x # 3 s !n 0.824 3 0.042 !200 0.824 0.0089 inch
6.8. In measuring reaction time, a psychologist estimates that the standard deviation is 0.05 second. How large a sample of measurements must he take in order to be (a) 95%, (b) 99% confident that the error in his estimate of mean reaction time will not exceed 0.01 second # (a) The 95% confidence limits are X 1.96s> !n, the error of the estimate being 1.96s> !n. Taking s s 0.05 second, we see that this error will be equal to 0.01 second if (1.96)(0.05)> !n 0.01, i.e., !n (1.96)(0.05)>0.01 9.8, or n 96.04. Therefore, we can be 95% confident that the error in the estimate will be less than 0.01 if n is 97 or larger. # (b) The 99% confidence limits are X 2.58s> !n. Then (2.58)(0.05) > !n 0.01, or n 166.4. Therefore, we can be 99% confident that the error in the estimate will be less than 0.01 only if n if 167 or larger. Note that the above solution assumes a nearly normal distribution for X, which is justified since the n obtained is large. 6.9. A random sample of 50 mathematics grades out of a total of 200 showed a mean of 75 and a standard deviation of 10. (a) What are the 95% confidence limits for the mean of the 200 grades (b) With what degree of confidence could we say that the mean of all 200 grades is 75 1 (a) Since the population size is not very large compared with the sample size, we must adjust for sampling without replacement. Then the 95% confidence limits are # X 1.96sX # X 1.96 s N AN !n n 1 75 1.96 10 200 A 200 !50 50 1 75 2.4 (b) The confidence limits can be represented by # X zcsX # X zc s N AN !n n 1 75 zc 200 A 200 !50 (10) 50 1 75 1.23zc 0 Since this must equal 75 1, we have 1.23zc 1 or zc 0.81. The area under the normal curve from z to z 0.81 is 0.2910; hence, the required degree of confidence is 2(0.2919) 0.582 or 58.2%. CHAPTER 6 Estimation Theory
Confidence interval estimates for means (small samples) 6.10. The 95% critical values (twotailed) for the normal distribution are given by 1.96. What are the corresponding values for the t distribution if the number of degrees of freedom is (a) n 9, (b) n 20, (c) n 30, (d) n 60 For 95% critical values (twotailed) the total shaded area in Fig. 63 must be 0.05. Therefore, the shaded area in the right tail is 0.025, and the corresponding critical value is t0.975. Then the required critical values are t0.975. For the given values of n these are (a) 2.26, (b) 2.09, (c) 2.04, (d) 2.00. Fig. 63 6.11. A sample of 10 measurements of the diameter of a sphere gave a mean x 4.38 inches and a standard deviation s 0.06 inch. Find (a) 95%, (b) 99% confidence limits for the actual diameter. # (a) The 95% confidence limits are given by X t0.975(S> !n 1). Since n n 1 10 1 9, we find t0.975 2.26 [see also Problem 6.10(a)]. Then using x 4.38 and s 0.06, the required 95% confidence limits are # 4.38 2.26 0.06 !10 1 4.38 0.0452 inch 0.045 4.335 inches and Therefore, we can be 95% confident that the true mean lies between 4.38 4.38 0.045 4.425 inches. (b) For n 9, t0.995 # X 3.25. Then the 99% confidence limits are t0.995(S> !n 1) 4.38 3.25(0.06> !10 1) 4.38

