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Answer :
To find the 99% confidence interval for the mean high temperature of the small towns, we will follow these steps:\n\n1. Identify the Sample Mean ([tex]\bar{x}[/tex]) and Sample Standard Deviation ([tex]s[/tex]):\n \n The high temperatures provided are: 98.4, 97.2, 96.7, 97.9, 96.3, 99.4, 98.9, 98.7, 96.4.\n \n First, calculate the sample mean ([tex]\bar{x}[/tex]):\n \n [tex]\bar{x} = \frac{98.4 + 97.2 + 96.7 + 97.9 + 96.3 + 99.4 + 98.9 + 98.7 + 96.4}{9} = \frac{880.9}{9} \approx 97.88[/tex]\n \n Next, calculate the sample standard deviation ([tex]s[/tex]):\n \n [tex]s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}[/tex]\n \n [tex]s = \sqrt{\frac{(98.4-97.88)^2 + (97.2-97.88)^2 + \dots + (96.4-97.88)^2}{8}} \approx 1.058[/tex]\n\n2. Determine the Critical Value (( t^ )) for a 99% Confidence Interval:\n \n Since the sample size is 9 (which leads to 8 degrees of freedom), and we want a 99% confidence interval, we look at the t-distribution table for [tex]df = 8[/tex]. The critical value ( t^ ) for a two-tailed test at the 99% confidence level is approximately 3.355.\n\n3. Calculate the Margin of Error (ME):\n \n [tex]ME = t^* \times \frac{s}{\sqrt{n}}[/tex]\n \n [tex]ME = 3.355 \times \frac{1.058}{\sqrt{9}} = 3.355 \times 0.3527 \approx 1.183[/tex]\n\n4. Construct the Confidence Interval:\n \n [tex]\bar{x} \pm ME = 97.88 \pm 1.183[/tex]\n \n Thus, the 99% confidence interval for the mean high temperature is approximately [tex](96.697, 99.063)[/tex].\n\nThis means we are 99% confident that the true mean high temperature of these small towns lies between 96.697°F and 99.063°F.
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