Question

webwork / math243spring-ozbek / week_8b_-_ch20_inference_about_a_population mean/5 Week 8b - Ch20 Inference About a Population Mean: Problem 5 Previous Problem List Next (1 point) Our bodies have a natural electrical field that is known to help wounds heal. Does changing the field strength slow healing? A series of experiments with newts investigated this question. The data below are the healing rates of cuts (micrometers per hour) in a matched pairs experiment. The pairs are the two hind limbs of the same newt, with the body's natural electric field in one limb (control) and half the natural electric value in the other limb (experimental). We're interested in knowing whether canceling the body's electric field reduces the healing rate. Give a 90% confidence interval for the difference in healing rates (control minus experimental). Newt 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Control 25 13 44 45 57 42 50 36 35 38 43 31 26 48 Experimental 24 23 47 42 26 46 8 28 2127 25 45 to Note: You can earn partial credit on this problem. Preview My Answers Submit Answers You have attempted this problem 0 times. You have unlimited attempts remaining. w d) earch O Et

Answer 1

$\begin{tabular}{rrr} \hline X & X - mean & (X-mean)\^{}2 \\ \hline 25 & -13.0714 & 170.862 \\ 13 & -25.0714 & 628.575 \\ 44 & 5.9286 & 35.1483 \\ 45 & 6.9286 & 48.0055 \\ 57 & 18.9286 & 358.292 \\ 42 & 3.9286 & 15.4339 \\ 50 & 11.9286 & 142.292 \\ 36 & -2.0714 & 4.2907 \\ 35 & -3.0714 & 9.4335 \\ 38 & -0.0714 & 0.0051 \\ 43 & 4.9286 & 24.2911 \\ 31 & -7.0714 & 50.0047 \\ 26 & -12.0714 & 145.719 \\ 48 & 9.9286 & 98.5771 \\ \hline \end{tabular}$

Confidence interval(in %) = 90

$\begin{tabular}{rrr} \hline X & X - mean & (X-mean)\^{}2 \\ \hline 24 & -8.3571 & 69.8411 \\ 23 & -9.3571 & 87.5553 \\ 47 & 14.6429 & 214.415 \\ 42 & 9.6429 & 92.9855 \\ 26 & -6.3571 & 40.4127 \\ 46 & 13.6429 & 186.129 \\ 38 & 5.6429 & 31.8423 \\ 33 & 0.6429 & 0.4133 \\ 28 & -4.3571 & 18.9843 \\ 28 & -4.3571 & 18.9843 \\ 21 & -11.3571 & 128.984 \\ 27 & -5.3571 & 28.6985 \\ 25 & -7.3571 & 54.1269 \\ 45 & 12.6429 & 159.843 \\ \hline \end{tabular} \\\sum_{i=1}^n x_i = 453.0 \\and\ n = 14 \\This\ implies\ that \\Mean(\bar{x}) = \frac{453.0}{14} \\Mean(\bar{x}) = 32.3571 \\(\sum{x_i - \bar{x}})^2 = 1133.214 \\n = 14 \\Variance(s^2) = \frac{1133.214}{13} \\Variance(s^2) = 87.1703 \\Standard\;Deviation(s) = \sqrt{Variance}$

Standard Deviation(s) = 9.3365 Mean 1(X1) = 38.0714 Sample size 1(ni) = 14 Standard deviation 1(si) = 11.539 Mean 2(X2) = 32.3571 Sample size 2(n2) = 14 Standard deviation 2(S2) = 9.3365 ta/2,01 +12-2 1.7033 Since we know that Confidence interval = Xi – X2=ta/2,1–1SpV (n1 – 1)s12 + (n2 – 1) s22 Sp = ni + n2 - 2 1 1 + ni n2 1 Required confidence interval (38.0714–32.3571–1.7033Spy i 1 14 38.0714- 9 32.3571 +1.7033SPV + 14 14

Required confidence interval = (5.7143-6.757, 5.7143+6.757)

Required confidence interval = (-1.0427, 12.4713)