document.write( "Question 1152873: A random sample of 16 pharmacy customers showed the waiting times below (in minutes). Find a 90 percent confidence interval for μ, assuming that the sample is from a normal population.
\n" ); document.write( "15 19 16 18 10 20 19 13
\n" ); document.write( "22 9 19 22 18 20 24 19
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Algebra.Com's Answer #774987 by jim_thompson5910(35256)\"\" \"About 
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\n" ); document.write( "Given Data Set = {15,19,16,18,10,20,19,13,22,9,19,22,18,20,24,19}
\n" ); document.write( "n = 16
\n" ); document.write( "because n > 30 is not true, and because we don't know sigma (population standard deviation), we use a T distribution
\n" ); document.write( "df = n-1 = 16-1 = 15 degrees of freedom\r
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\n" ); document.write( "\n" ); document.write( "Use a calculator or spreadsheet program to find these two values
\n" ); document.write( "xbar = sample mean = 17.6875
\n" ); document.write( "s = sample standard deviation = 4.1748253456 which is approximate\r
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\n" ); document.write( "\n" ); document.write( "In the back of your stats textbook is a t table that looks similar to this shown here
\n" ); document.write( "http://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
\n" ); document.write( "Using that table specifically, highlight the entire row that starts with df = 15. Highlight the column that represents the 90% confidence level (see the bottom portion of the table).
\n" ); document.write( "The value 1.753 is in this row and column. Therefore, t = 1.753 is the approximate critical value.
\n" ); document.write( "Note: P(|T| < 1.753) = 0.90 which is another way of saying P(-1.753 < T < 1.753) = 0.90\r
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\n" ); document.write( "\n" ); document.write( "Now onto computing the lower (L) and upper (U) bounds of the confidence interval.
\n" ); document.write( "L = lower bound of confidence interval
\n" ); document.write( "L = xbar - t*(s/sqrt(n))
\n" ); document.write( "L = 17.6875 - 1.753*(4.1748253456/sqrt(16))
\n" ); document.write( "L = 17.6875 - 1.753*(4.1748253456/4)
\n" ); document.write( "L = 17.6875 - 1.753*(1.0437063364)
\n" ); document.write( "L = 17.6875 - 1.8296172077092
\n" ); document.write( "L = 15.8578827922908
\n" ); document.write( "L = 15.86\r
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\n" ); document.write( "\n" ); document.write( "U = upper bound of confidence interval
\n" ); document.write( "U = xbar + t*(s/sqrt(n))
\n" ); document.write( "U = 17.6875 + 1.753*(4.1748253456/sqrt(16))
\n" ); document.write( "U = 17.6875 + 1.753*(4.1748253456/4)
\n" ); document.write( "U = 17.6875 + 1.753*(1.0437063364)
\n" ); document.write( "U = 17.6875 + 1.8296172077092
\n" ); document.write( "U = 19.5171172077092
\n" ); document.write( "U = 19.52\r
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\n" ); document.write( "\n" ); document.write( "(L,U) = (15.86,19.52)\r
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\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "The 90% confidence interval is approximately (15.86,19.52)\r
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\n" ); document.write( "\n" ); document.write( "We can write the confidence interval in the form L < mu < U which means we would have 15.86 < mu < 19.52\r
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\n" ); document.write( "\n" ); document.write( "Yet another alternative is to write the confidence interval in the form (MoE stands for Margin of Error), so we would get when rounding to 2 decimal places.
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