document.write( "Question 922869: A sample of 10 observations is selected from a normal population for which the population standard deviation is known to be 3. The sample mean is 23. (Round your answers to 3 decimal places.)
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\n" ); document.write( "\n" ); document.write( "(a) The standard error of the mean is ....\r
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\n" ); document.write( "I have already solved (a) and the standard error of the mean is .948 but I am having trouble with the confidence interval. Can you please show your work so I can understand how to work these types of problems? \n" ); document.write( "
Algebra.Com's Answer #559773 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
 
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document.write( "Hi
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document.write( "mean = 23, sd(population) = 3
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document.write( "a) SE = sd/sqrt(n) = 3/sqrt(10) = .9487
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document.write( "b) 99% CI (see chart below)
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document.write( "ME = 3.25SE = 3.25(.9487) = 3.08 = 3 
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document.write( "Round the confidence interval limits to the same number of decimal places as the sample mean
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document.write( "ME = 3
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document.write( "CI:  23-ME < u < 23 + ME
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document.write( "CI: (20, 26)
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document.write( ".......Small Samples
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document.write( "DF	Probability, p 2-tailed			
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document.write( "	0.1	0.05	0.01	0.001
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document.write( "1	6.31	12.71	63.66	636.62
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document.write( "2	2.92	4.3	9.93	31.6
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document.write( "3	2.35	3.18	5.84	12.92
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document.write( "4	2.13	2.78	4.6	8.61
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document.write( "5	2.02	2.57	4.03	6.87
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document.write( "6	1.94	2.45	3.71	5.96
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document.write( "7	1.89	2.37	3.5	5.41
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document.write( "8	1.86	2.31	3.36	5.04
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document.write( "9	1.83	2.26	\"+highlight_green%283.25%29\" 	4.78
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document.write( "10	1.81	2.23	3.17	4.59
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document.write( "....Larger Samples
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document.write( " = CI	z = value
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document.write( "90%	z =1.645
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document.write( "92%	z = 1.751
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document.write( "95%	z = 1.96
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document.write( "98%	z = 2.326
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document.write( "99%	z = 2.576 \r
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