document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1185509>Question 1185509</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 11 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with 𝜎 = 0.36 gram.<BR>\n" );
document.write( "When finding an 80% confidence interval, what is the critical value for confidence level? (Give your answer to two decimal places.)<BR>\n" );
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document.write( "Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.)<BR>\n" );
document.write( "lower limit	<BR>\n" );
document.write( "upper limit	<BR>\n" );
document.write( "margin of error	<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849691 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1987)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=849691\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's how to calculate the 80% confidence interval and margin of error:\r<BR>\n" );
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document.write( "1. **Find the critical value (z_c):**\r<BR>\n" );
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document.write( "For an 80% confidence level, the alpha (α) is 1 - 0.80 = 0.20. Since confidence intervals are two-tailed, we divide alpha by 2: 0.20 / 2 = 0.10.\r<BR>\n" );
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document.write( "We want to find the z-score that corresponds to an area of 0.10 in *each* tail of the standard normal distribution. This means the area to the *left* of the z-score is 1 - 0.10 = 0.90.  Using a standard normal (z) table or calculator, look up the z-score corresponding to a cumulative area of 0.90.  You'll find that z_c ≈ 1.28.\r<BR>\n" );
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document.write( "So, z_c = 1.28\r<BR>\n" );
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document.write( "2. **Calculate the margin of error (E):**\r<BR>\n" );
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document.write( "The margin of error is calculated as:\r<BR>\n" );
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document.write( "E = z_c * (σ / √n)\r<BR>\n" );
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document.write( "Where:\r<BR>\n" );
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document.write( "* z_c is the critical value (1.28)<BR>\n" );
document.write( "* σ is the population standard deviation (0.36 gram)<BR>\n" );
document.write( "* n is the sample size (11 hummingbirds)\r<BR>\n" );
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document.write( "E = 1.28 * (0.36 / √11)<BR>\n" );
document.write( "E ≈ 1.28 * (0.36 / 3.3166)<BR>\n" );
document.write( "E ≈ 1.28 * 0.1085<BR>\n" );
document.write( "E ≈ 0.14 gram (rounded to two decimal places)\r<BR>\n" );
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document.write( "3. **Calculate the confidence interval:**\r<BR>\n" );
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document.write( "The confidence interval is calculated as:\r<BR>\n" );
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document.write( "(x̄ - E, x̄ + E)\r<BR>\n" );
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document.write( "Where x̄ is the sample mean (3.15 grams).\r<BR>\n" );
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document.write( "Lower Limit = x̄ - E = 3.15 - 0.14 = 3.01 grams<BR>\n" );
document.write( "Upper Limit = x̄ + E = 3.15 + 0.14 = 3.29 grams\r<BR>\n" );
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document.write( "Therefore:\r<BR>\n" );
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document.write( "* z_c = 1.28<BR>\n" );
document.write( "* Lower limit = 3.01 grams<BR>\n" );
document.write( "* Upper limit = 3.29 grams<BR>\n" );
document.write( "* Margin of error = 0.14 grams<BR>\n" );
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