Question

(1 point) Archeologists have discovered a sample of ancient Egyptian skulls from two different time periods, 4000 B.C. and 150 A.D. They believe that the mean maximum skull breadth (the width of the skull at its widest part) has increased from 4000 B.C. to 150 A.C.A sample of 13 Egyptian skull breadths (in mm) from each time period is show below: 4000 B.C. 118, 125, 134, 129, 122, 129, 132, 129 150 A.D. 132, 131, 137, 140, 142, 134, 132, 128 - Part 1: Calculate the mean and standard deviation for each data set. Round each mean to 2 decimal places, and each standard deviation to 4 decimal places. 4000 B.C. mean: 127.25 150 A.D. mean: 134.50 4000 B.C. standard deviation: 5.2847 150 A.D. standard deviation: 4.7809 Part 2: Using your answers from part 1 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 VX where x is any number, type sqrt(x). For example, V2 should be written as sqrt(2). 4000 B.C. skull breadths: to s

Answer 1

Let 4000 BC = X

150 AD = Y

	X	Y
	118	132
	125	131
	134	137
	129	140
	122	142
	129	134
	132	132
	129	128
MEAN &y	127.2500	134.5000
S.D	5.2847	4.7809

95% Confidence Interval of is

1

$\bar{x} - \frac{s_{x}}{\sqrt{n-1}}*t_{cri} \leq\mu _{x} \leq \bar{x} + \frac{s_{x}}{\sqrt{n-1}}*t_{cri}$

Since Confidence Interval = 95%; So, Level of Significance = $\alpha$ = 5% = 0.05

n = Number of Observations = 8

$\Rightarrow t_{cri} = t_{(n-1),\alpha } = t_{(8-1),5\%}=t_{7,0.05} = 2.365$

Therefore the

95% Confidence Interval of is

1

$\bar{x} - \frac{s_{x}}{\sqrt{n-1}}*t_{cri} \leq\mu _{x} \leq \bar{x} + \frac{s_{x}}{\sqrt{n-1}}*t_{cri}$

$\Rightarrow 127.25 - \frac{5.2847}{\sqrt{8-1}}*2.365 \leq \mu _{x} \leq 127.25 + \frac{5.2847}{\sqrt{8-1}}*2.365$

$\Rightarrow 127.25 - 1.9974*2.365 \leq \mu _{x} \leq 127.25 +1.9974*2.365$

$\Rightarrow 127.25 - 4.7239 \leq \mu _{x} \leq 127.25 + 4.7239$

$\Rightarrow 122.5261 \leq \mu _{x} \leq 131.9739$

Similarly the 95% Confidence Interval of $\mu _{y}$ is

$\bar{y} - \frac{s_{y}}{\sqrt{n-1}}*t_{cri} \leq\mu _{y} \leq \bar{y} + \frac{s_{y}}{\sqrt{n-1}}*t_{cri}$

Since Confidence Interval = 95%; So, Level of Significance = $\alpha$ = 5% = 0.05

n = Number of Observations = 8

$\Rightarrow t_{cri} = t_{(n-1),\alpha } = t_{(8-1),5\%}=t_{7,0.05} = 2.365$

Therefore the

95% Confidence Interval of $\mu _{y}$ is

$\bar{y} - \frac{s_{y}}{\sqrt{n-1}}*t_{cri} \leq\mu _{y} \leq \bar{y} + \frac{s_{y}}{\sqrt{n-1}}*t_{cri}$

$\Rightarrow 134.5 - \frac{4.7809}{\sqrt{8-1}}*2.365 \leq \mu _{y} \leq 134.5 + \frac{4.7809}{\sqrt{8-1}}*2.365$

$\Rightarrow134.5 - 1.8074*2.365 \leq \mu _{y} \leq 134.5 + 1.8074*2.365$

$\Rightarrow134.5 - 4.2736 \leq \mu _{y} \leq 134.5 + 4.2736$

$\Rightarrow130.2264 \leq \mu _{y} \leq 138.7736$

TABULATED VALUE OF t:

t 999 t Table cum. prob. one-tail two-tails df t.50 0.50 1.00 0.25 0.50 7.80 0.20 0.40 1.8 0.15 0.30 t90 0.10 0.20 4.95 0.05 0.10 t 975 0.025 0.05 t 99 0.01 0.02 t 995 t .9995 0.005 0.001 0.0005 0.01 0.002 0.001 1.376 1.061 0.978 0.941 0.920 5.893 6.869 2 31 41 5 6 71 81 9 100 11 121 1.356 1.782 3.930 13 14 15 16 171 18 19 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0% 1.000 0.816 0.765 0.741 0.727 0.718 0.711 0.706 0.703 0.700 0.697 0.695 0.694 0.692 0.691 0.690 0.689 0.688 0.688 0.687 0.686 0.686 0.685 0.685 0.684 0.684 0.684 0.683 0.683 0.683 0.681 0.679 0.678 0.677 0.675 0.674 1.963 1.386 1.250 1.190 1.156 1.134 1.119 1.108 1.100 1.093 1.088 1.083 1.079 1.076 1.074 1.071 1.069 1.067 1.066 1.064 1.063 1.061 1.060 1.059 1.058 1.058 1.057 1.056 1.055 1.055 1.050 1.045 1.043 1.042 1.037 0.906 0.896 0.889 0.883 0.879 0.876 0.873 0.870 0.868 0.866 0.865 0.863 0.862 0.861 0.860 0.859 0.858 0.858 0.857 0.856 0.856 0.855 0.855 0.854 0.854 0.851 0.848 0.846 0.845 0.842 0.842 3.078 6.314 12.71 1.886 2.920 4.303 1.638 2.353 3.182 1.533 2.132 2.776 1.476 2.015 2.571 1.440 1.943 2.447 1.415 1.895 2.365 1.397 1.860 2.306 1.383 1.833 2.262 1.372 1.812 2.228 1.363 1.796 2.201 2.179 1.350 1.771 2.160 1.345 1.761 2.145 1.341 1.753 2.131 1.337 1.746 2.120 1.333 1.740 2.110 1.330 1.734 2.101 1.328 1.729 2.093 1.325 1.725 2.086 1.323 1.721 2.080 1.321 1.717 2.074 1.319 1.714 2.069 1.318 1.711 2.064 1.316 1.708 2.060 1.315 1.706 2.056 1.314 1.703 2.052 1.313 1.701 2.048 1.311 1.699 2.045 1.310 1.697 2.042 1.303 1.684 2.021 1.296 1.671 2.000 1.292 1.664 1.990 1.290 1.660 1.984 1.282 1.646 1.962 1.282 1.645 1.960 80% 90% 95% Confidence Level 31.82 6.965 4.541 3.747 3.365 3.143 2.998 2.896 2.821 2.764 2.718 2.681 2.650 2.624 2.602 2.583 2.567 2.552 2.539 2.528 2.518 2.508 2.500 2.492 2.485 2.479 2.473 2.467 2.462 2.457 2.423 2.390 2.374 2.364 2.330 2.326 98% 20 63.66 318.31 636.62 9.925 22.327 31.599 5.841 10.215 12.924 4.604 7.173 8.610 4.032 3.707 5.208 5.959 3.499 4.785 5.408 3.355 4.501 5.041 3.250 4.297 4.781 3.169 4.144 4.587 3.106 4.025 4.437 3.055 4.318 3.012 3.852 4.221 2.977 3.787 4.140 2.947 3.733 4.073 2.921 3.686 4.015 2.898 3.646 3.965 2.878 3.610 3.922 2.861 3.579 3.883 2.845 3.552 3.850 2.831 3.527 3.819 2.819 3.792 2.807 3.485 3.768 2.797 3.467 3.745 2.787 3.450 3.725 2.779 3.435 3.707 2.771 3.421 3.690 2.763 3.408 3.674 2.756 3.396 3.659 2.750 3.385 3.646 2.704 3.307 3.551 2.660 3.232 3.460 2.639 3.195 3.416 2.626 3.174 3.390 2.581 3.098 3.300 2.576 3.090 3.291 99% 99.8% 99.9% 21 22 23 24 25 26 27 28 29 30 40 601 80 100 3.505 1000 Z 1.036 70% 50% 60% t-table.xls 7/14/2007

IIn this Search the Critical value of t; which is intersected between 7 and 0.05 (Two tails ); we will get 2.365