document.write( "Question 1175140: 1. You measure 33 backpacks' weights, and find they have a mean weight of 44 ounces. Assume the population standard deviation is 12.1 ounces. Based on this, what is the maximal margin of error associated with a 90% confidence interval for the true population mean backpack weight.Give your answer as a decimal, to two places.\r
\n" ); document.write( "\n" ); document.write( "2. You intend to estimate a population mean μ with the following sample.
\n" ); document.write( "80.4
\n" ); document.write( "78.8
\n" ); document.write( "68.6
\n" ); document.write( "52.4
\n" ); document.write( "59.4
\n" ); document.write( "63.8
\n" ); document.write( "61.2
\n" ); document.write( "67.4
\n" ); document.write( "63
\n" ); document.write( "72.1
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\n" ); document.write( "77.8\r
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\n" ); document.write( "\n" ); document.write( "You believe the population is normally distributed. Find the 80% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to twp decimal places (because the sample data are reported accurate to one decimal place).
\n" ); document.write( "80% C.I. =
\n" ); document.write( "Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places. \n" ); document.write( "

Algebra.Com's Answer #800730 by ewatrrr(24785) $\"\"$ $\"About$

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document.write( "Hi\r\n" );
document.write( "1) Normal Distribution: the population standard deviation σ  = 12.1 ounces \r\n" );
document.write( "   Sample Size is 33: Sample mean = 44  \r\n" );
document.write( "90% confidence Interval, find ME \r\n" );
document.write( "   ME = 1.645 (12.1)/√33 = 3.465\r\n" );
document.write( "CI = 44 ± 3.465\r\n" );
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document.write( "2)  Normal Distribution:  Sample of 12    mean = 66.533   sd = 9.012  \r\n" );
document.write( "80% CI  z = 1.362 df 11\r\n" );
document.write( "ME == = 3.543\r\n" );
document.write( "80% C.I. = (62.99,70.08)\r\n" );
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document.write( "Wish You the Best in your Studies.\r\n" );
document.write( "Confidence Level	80%	90%	95%	98%	99%\r\n" );
document.write( "Two-sided test p-values	0.2	0.1	0.05	0.02	0.01\r\n" );
document.write( "One-sided test p-values	0.1	0.05	0.025	0.01	0.005\r\n" );
document.write( "Degrees of Freedom (df=n-1)					\r\n" );
document.write( "               1	3.078	6.314	12.71	31.82	63.66\r\n" );
document.write( "               2	1.886	2.92	4.303	6.965	9.925\r\n" );
document.write( "               3	1.638	2.353	3.182	4.541	5.841\r\n" );
document.write( "               4	1.533	2.132	2.776	3.747	4.604\r\n" );
document.write( "               5	1.476	2.015	2.571	3.365	4.032\r\n" );
document.write( "               6	1.44	1.943	2.447	3.143	3.707\r\n" );
document.write( "               7	1.415	1.895	2.365	2.998	3.499\r\n" );
document.write( "               8	1.397	1.86	2.306	2.896	3.355\r\n" );
document.write( "               9	1.383	1.833	2.262	2.821	3.25\r\n" );
document.write( "              10	1.372	1.812	2.228	2.764	3.169\r\n" );
document.write( "              11	1.362	1.796	2.201	2.718	3.106\r\n" );
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