document.write( "Question 1173384: Question :
\n" ); document.write( "A sample of 49 is drawn from a normal population with unknown mean and variance. If the sample means is 131.8 and standard deviation 6.25,setup a 90% confidence interval for the population mean?\r
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Algebra.Com's Answer #798765 by Theo(13342)\"\" \"About 
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sample size is 49
\n" ); document.write( "mean of sample is 131.8
\n" ); document.write( "standard deviation of sample is 6.25\r
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\n" ); document.write( "\n" ); document.write( "you want to set up a 90% confidence interval.\r
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\n" ); document.write( "\n" ); document.write( "degrees of freedom is 49 - 1 = 48.\r
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\n" ); document.write( "\n" ); document.write( "standard error = standard deviation of sample divided by square root of sample size.\r
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\n" ); document.write( "\n" ); document.write( "you will be using t-scores with 48 degrees of freedom.\r
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\n" ); document.write( "\n" ); document.write( "standard error = 6.25 / sqrt(49) = 6.25 / 7 = .8928571429.
\n" ); document.write( "i rounded this to .892857 which is probability a little more than close enouogh.\r
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\n" ); document.write( "\n" ); document.write( "in the t-score table, with 90% confidence interval, one tailed alpha would be 5% and two tailed alpha would be 10% (5% on each end).\r
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\n" ); document.write( "\n" ); document.write( "here's the table i used.\r
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\n" ); document.write( "\n" ); document.write( "https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf\r
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\n" ); document.write( "\n" ); document.write( "that table only showed degrees of freedom of 40 and 60.
\n" ); document.write( "48 degrees of freedom is somewhere in between.\r
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\n" ); document.write( "\n" ); document.write( "fortunately, there is a t-score calculator online that can be used to get more accurate critical t-score.\r
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\n" ); document.write( "\n" ); document.write( "that t-score calculator can be found at https://stattrek.com/online-calculator/t-distribution.aspx\r
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\n" ); document.write( "\n" ); document.write( "the calculator gets you the area to the left of the t-score.
\n" ); document.write( "to find the two tailed 10% conidence interval, i entered .05 in the p(T < t) box and a mean of 0 and a standard deviatiion of 1 and degrees of freedom of 48 and clicked on calculate.\r
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\n" ); document.write( "\n" ); document.write( "calculator said the t-score was -1.677
\n" ); document.write( "since the normal distribution table is symmetric, my confidence interval would be between a t-score of -1.677 and a t-score of 1.677
\n" ); document.write( "that would be 5% area to the left of -1.677 and 5% area to the right of -1.677.\r
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\n" ); document.write( "\n" ); document.write( "in the table, two tailed of 10% with 40 degrees of freedom gave a t-score of -1.684 and 60 degrees of freedom gave a t-score of 1.671.
\n" ); document.write( "about halfway between those t-score would yield a t-score of -1.6775 which is pretty close to -1.677 found by the calculator.\r
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\n" ); document.write( "\n" ); document.write( "i used a critical t-score of plus or minus 1.677 for my 10% two tailed confidence interval.\r
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\n" ); document.write( "\n" ); document.write( "once this is done, you need to use the t-score formula to find the raw score.\r
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\n" ); document.write( "\n" ); document.write( "t-score formula is t = (x - m) / s
\n" ); document.write( "t is the t-score
\n" ); document.write( "x is the raw score
\n" ); document.write( "m is the mean
\n" ); document.write( "s is the standard error.\r
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\n" ); document.write( "\n" ); document.write( "for t = -1.677, i got:
\n" ); document.write( "-1.677 = (x - 131.8) / .892857
\n" ); document.write( "solve for x to get:
\n" ); document.write( "x = -1.677 * .892857 + 131.8 = 130.3026788\r
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\n" ); document.write( "\n" ); document.write( "for t = 1.677, i got:
\n" ); document.write( "1.677 = (x - 131.8) / .892857
\n" ); document.write( "solve for x to get:
\n" ); document.write( "x = 1.677 * .892857 + 131.8 = 133.2973212\r
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\n" ); document.write( "\n" ); document.write( "your answer should be:\r
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\n" ); document.write( "\n" ); document.write( "90% confidence interval is between 130.3026788 and 133.2973212.\r
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