document.write( "Question 1193902: A publisher wants to estimate the mean length of time (in minutes) all adults spend reading newspapers. To determine this estimate, the publisher takes a random sample of 15 people and obtains the results below. From past studies, the publisher assumes that the population of times is normally distributed. Construct the 95% confidence interval for the population mean. \r
\n" ); document.write( "\n" ); document.write( "12 8 11 10 11 6 6 9 6 11 11 11 7 7 9
\n" ); document.write( "The sample mean is
\n" ); document.write( "The sample standard deviation rounded to 2 decimal places is
\n" ); document.write( "The number of degrees of freedom is
\n" ); document.write( "According to the t-distribution table in the formula sheets t c rounded to 3 decimal places is
\n" ); document.write( "The margin of error E rounded to two decimal places is
\n" ); document.write( "The confidence interval is < 𝜇 <
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Algebra.Com's Answer #825990 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "There are n = 15 values in the list, which is the sample size.\r
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\n" ); document.write( "\n" ); document.write( "To get the sample mean, first we add up the values
\n" ); document.write( "12+8+11+10+11+6+6+9+6+11+11+11+7+7+9 = 135
\n" ); document.write( "Then divide this over the sample size (n = 15) to get
\n" ); document.write( "135/15 = 9
\n" ); document.write( "The sample mean is 9
\n" ); document.write( "We call this xbar because x has a horizontal bar over top.
\n" ); document.write( "xbar = 9\r
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\n" ); document.write( "\n" ); document.write( "You could calculate the sample standard deviation by hand, but it's preferable to use technology instead. There are many free calculators online to help with that if you don't have a TI83/84.
\n" ); document.write( "You should get a sample standard deviation of roughly s = 2.17
\n" ); document.write( "The sample standard deviation s estimates the population standard deviation sigma.
\n" ); document.write( "s = sample standard deviation = 2.17 approximately
\n" ); document.write( "sigma = population standard deviation = unknown\r
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\n" ); document.write( "\n" ); document.write( "Because we don't know sigma and because n > 30 is not true, we must use a T distribution.
\n" ); document.write( "The degrees of freedom (df) is equal to the sample size minus 1
\n" ); document.write( "df = n-1
\n" ); document.write( "df = 15-1
\n" ); document.write( "df = 14\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Use the T table in the back of your textbook, or use a free online resource like this
\n" ); document.write( "https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
\n" ); document.write( "Start by looking at the row labeled \"two tails\". The value 0.05 corresponds to a confidence interval of 95% since 1 - 0.05 = 0.95
\n" ); document.write( "The 0.05 refers to the combined area of both tails.
\n" ); document.write( "Highlight this entire column.
\n" ); document.write( "Then look in the row that has df = 14 at the very left.
\n" ); document.write( "The intersection of the row and column leads to the approximate t critical value of 2.145
\n" ); document.write( "What is this value saying? It says that P(-2.145 < T < 2.145) = 0.95 approximately when df = 14.
\n" ); document.write( "95% of the area under the T curve (df = 14) is between roughly t = -2.145 and t = 2.145\r
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\n" ); document.write( "\n" ); document.write( "Let's compute the margin of error
\n" ); document.write( "\"E+=+t%5Bc%5D%2Aexpr%28s%2Fsqrt%28n%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"E+=++2.145%2Aexpr%282.17%2Fsqrt%2815%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"E+=++1.20182546216162\"\r
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\n" ); document.write( "\n" ); document.write( "\"E+=++1.20\"\r
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\n" ); document.write( "\n" ); document.write( "And lastly, let's compute the confidence interval for the mean mu (symbol 𝜇)
\n" ); document.write( "\"xbar+-+E+%3C+mu+%3C+xbar+%2B+E\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"9+-+1.20+%3C+mu+%3C+9+%2B+1.20\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"7.80+%3C+mu+%3C+10.20\"
\n" ); document.write( "We could shorten this to (7.80, 10.20) which is common notation for confidence intervals.
\n" ); document.write( "Some books and research papers use the notation [7.80, 10.20] to mean the same thing.
\n" ); document.write( "We are 95% confident the true population mean (mu) is somewhere between 7.80 and 10.20\r
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\n" ); document.write( "\n" ); document.write( "------------------------------------\r
\n" ); document.write( "\n" ); document.write( "Answers:
\n" ); document.write( "Sample mean = 9
\n" ); document.write( "Sample standard deviation = 2.17
\n" ); document.write( "Degrees of Freedom = 14
\n" ); document.write( "\"t%5Bc%5D\" (t critical value) is roughly 2.145
\n" ); document.write( "Margin of error: E = 1.20
\n" ); document.write( "The 95% confidence interval is roughly \"7.80+%3C+mu+%3C+10.20\"
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