Question

(1 point) Archeologists have discovered a sample of ancient Egyptian skulls from 150 A.C. A sample of 5 Egyptian skull breadths (in mm) from 150 A.C. is show below: Skull breadth Sample: 128, 128, 142, 139, 135 (a) Calculate the mean and standard deviation for for the data set. Round the mean to 2 decimal places, and the standard deviation to 4 decimal places. 150 A.C. skull breadths mean: 150 A.C. skull breadths standard deviation: (b) Using your answers from part (a) and the formula for a 95% confidence interval as presented in lecture, fill in the blanks with the appropriate values for this problem for calculating the confidence interval below. To enter V where x is any number, type sqrt(x). For example, V2 should be written as sqrt(2). 150 A.C. skull breadths: to (c) Using the formulas from part (b), find the 95% confidence intervals for the mean maximum skull breadths of all Egyptian skulls from 150 A.C. Round your answer to the nearest hundredth (2 decimal places). 150 A.C. skull breadths: to

Answer 1

= 134.4 n 1. Solution: Given data, x = (128, 128, 142, 139, 135) n = 5 128 + 128 + ... + 135 672 5 5 Lei (; -72 (128 - 134.4)2 + (128 – 134.4)2 + ... + (135 – 134.4)2 S= n - 1 (5 – 1) 161.2 6.3482 4 (a) 150 A.C. skull breadths mean: 134.4 150 A.C. skull breadths standard deviation: 6.3482 (6)The formula for 95% confidence interval is: 150 A.C. skull breadths: 1 – to,025,4 T1 to ã + to.025,4. To S (c) For 95% confidence interval +0.025,4 = 2.776 Therefore, 6.3482 6.3482 (134.4 - (2.776) ), 134.4+ (2.776) :)) (126.52, 142.28) 5 The 95% confidence interval for the mean maximum skull breadths of all ancient Egyptian skull from 150 A.C. fall between 126.52 to 142.28 5

$\\2.\: \textbf{\underline{Solution}:}\\\text{Given data,} \\x=\left (132,135,137,135,135 \right ) \\ n=5\\ \bar{x}=\frac{\sum_{i=1}^{n} x_i}{n}=\frac{132+135+...+135}{5}=\frac{674}{5}=134.8 \\ s=\sqrt{\frac{\sum_{i=1}^{n}(x_i-\bar{x})^2}{n-1}}=\sqrt{\frac{(132-134.8)^2+(135-134.8)^2+...+(135-134.8)^2}{(5-1)}}\\\:\:\: =\sqrt{\frac{12.8}{4}}\approx 1.7889 \\(a)\\\text{150 A.C. skull breadths mean: 134.8} \\\: \text{150 A.C. skull breadths standard deviation: 1.7889}$