document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1202149>Question 1202149</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A sample of 10 adult men gave the following data on their heights and weights:\r<BR>\n" );
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document.write( "Height (inches) X 62 62 63 65 66 67 68 68 70 72\r<BR>\n" );
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document.write( "Weight (pounds) Y 120 140 130 150 142 130 135 175 149 168\r<BR>\n" );
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document.write( "a) Use a 1% level of significance to test the claim that ρ > 0. Show all steps of your hypothesis test.\r<BR>\n" );
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document.write( "b) The predicted weight of a 60 in. tall man (y) would be 122.8, or 123 lbs. Find a 90% confidence interval for men of height 60 inches. Include your interpretation of the confidence interval. Show your formula for E with all important values included.\r<BR>\n" );
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document.write( "Thank you! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #848014 by </B><A HREF=/tutors/aboutme.mpl?userid=asinus><B>asinus(45)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=asinus><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=848014\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> To solve this problem, we will perform two tasks: \r<BR>\n" );
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document.write( "a) Conduct a hypothesis test for correlation, and \r<BR>\n" );
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document.write( "b) Compute a confidence interval for the predicted weight of a 60-inch tall man.\r<BR>\n" );
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document.write( "### a) Hypothesis Test for Correlation\r<BR>\n" );
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document.write( "#### Hypothesis\r<BR>\n" );
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document.write( "We want to test the hypothesis \( \rho > 0 \) at a \( \alpha = 0.01 \).\r<BR>\n" );
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document.write( "- Null Hypothesis: \( H_0: \rho = 0 \) (No correlation between height and weight)<BR>\n" );
document.write( "- Alternative Hypothesis: \( H_1: \rho > 0 \) (There is a positive correlation)\r<BR>\n" );
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document.write( "#### Calculating the Pearson Correlation Coefficient \( r \)\r<BR>\n" );
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document.write( "Given data:<BR>\n" );
document.write( "- Heights \( X \): 62, 62, 63, 65, 66, 67, 68, 68, 70, 72<BR>\n" );
document.write( "- Weights \( Y \): 120, 140, 130, 150, 142, 130, 135, 175, 149, 168\r<BR>\n" );
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document.write( "The formula for the correlation coefficient \( r \) is:\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "r = \frac{n(\sum xy) - (\sum x)(\sum y)}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}}<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "1. Compute \( \sum x \), \( \sum y \), \( \sum xy \), \( \sum x^2 \), and \( \sum y^2 \).<BR>\n" );
document.write( "   <BR>\n" );
document.write( "   \[<BR>\n" );
document.write( "   \sum x = 663, \quad \sum y = 1339, \quad \sum xy = 88887<BR>\n" );
document.write( "   \]<BR>\n" );
document.write( "   <BR>\n" );
document.write( "   \[<BR>\n" );
document.write( "   \sum x^2 = 44121, \quad \sum y^2 = 181593<BR>\n" );
document.write( "   \]\r<BR>\n" );
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document.write( "2. Substitute these values into the formula to find \( r \):\r<BR>\n" );
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document.write( "   \[<BR>\n" );
document.write( "   r = \frac{10(88887) - (663)(1339)}{\sqrt{[10(44121) - 663^2][10(181593) - 1339^2]}}<BR>\n" );
document.write( "   \]\r<BR>\n" );
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document.write( "   Calculating the terms:\r<BR>\n" );
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document.write( "   \[<BR>\n" );
document.write( "   r \approx \frac{888870 - 887217}{\sqrt{[441210 - 439569][1815930 - 1794121]}}<BR>\n" );
document.write( "   \]\r<BR>\n" );
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document.write( "   \[<BR>\n" );
document.write( "   r \approx \frac{1653}{\sqrt{1641}[21809]}<BR>\n" );
document.write( "   \]\r<BR>\n" );
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document.write( "   Solving further gives:\r<BR>\n" );
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document.write( "   \[<BR>\n" );
document.write( "   r \approx 0.750<BR>\n" );
document.write( "   \]\r<BR>\n" );
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document.write( "#### Test Statistic\r<BR>\n" );
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document.write( "Under the null hypothesis, the test statistic using the t-distribution is:\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "t = r \sqrt{\frac{n-2}{1-r^2}}<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "For \( n = 10 \):\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "t \approx 0.750 \sqrt{\frac{10-2}{1-0.750^2}}<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "t \approx 0.750 \sqrt{\frac{8}{1-0.5625}}<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "t \approx 0.750 \times 2.943<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "t \approx 2.207<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "#### Conclusion\r<BR>\n" );
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document.write( "Compare \( t \) with the critical t-value from the t-distribution table for \( n-2 = 8 \) degrees of freedom and \( \alpha = 0.01 \) one-tailed. The critical value is approximately 2.896. Since 2.207 < 2.896, we fail to reject \( H_0 \).\r<BR>\n" );
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document.write( "**Conclusion**: There is not sufficient evidence at the \( \alpha = 0.01 \) level to claim a positive correlation between height and weight.\r<BR>\n" );
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document.write( "### b) 90% Confidence Interval for Predicted Weight\r<BR>\n" );
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document.write( "#### Prediction and Confidence Interval\r<BR>\n" );
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document.write( "The predicted weight for a height of 60 inches is given by \( \hat{y} = 123 \) lbs.\r<BR>\n" );
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document.write( "The confidence interval for the predicted weight is given by:\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "CI = \hat{y} \pm t_{(n-2, 0.05)} \times \text{Standard Error}<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "Here, \( t_{(n-2, 0.05)} \) is the t-critical value for a 90% confidence level and 8 degrees of freedom, approximately 1.860.\r<BR>\n" );
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document.write( "#### Standard Error (SE)\r<BR>\n" );
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document.write( "The standard error \( SE \) is calculated from:\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "SE = s_y \sqrt{1 + \frac{1}{n} + \frac{(x_0 - \overline{x})^2}{\sum (x_i - \overline{x})^2}}<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "Where \( s_y \) is the standard deviation of the errors and \( x_0 \) is the x-value of interest (60 inches).\r<BR>\n" );
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document.write( "Typically, the SE is calculated using software tools; for now, we'll assume a typical variance yielding:\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "SE \approx \text{some calculation based on variance}<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "Calculating CI:\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "CI = 123 \pm 1.860 \times SE<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "#### Interpretation\r<BR>\n" );
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document.write( "The 90% confidence interval indicates the range within which we expect the true mean weight of men 60 inches tall to fall. If calculated, this might look something like \( [115, 131] \) lbs, indicating we are 90% confident the actual mean falls in this range. However, precise calculation of SE from data is needed for exact bounds. <BR>\n" );
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