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: 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 a where x is any number, type sqrt(x). For example, V2 should be written as sqrt(2). 4000 B.C. skull breadths: 5.2847 5.2847 to 127.25 2.3646 127.25 2.3646 sqrt(8) sqrt(8) 150 A.D. skull breadths: 4.7809 4.7809 to - 2.3646 134.5 134.5 + 2.3646 sqrt(8) sqrt(8) - Part 3 ...maximum skall hadithiwatin from

Answer 1

	4000 B.C	150 A.D
	118	132
	125	131
	134	137
	129	140
	122	142
	129	134
	132	132
	129	128
Mean	127.25	134.5	AVERAGE()
Sd	5.284749	4.780914	STDEV()
n	8	8

95% CI for 4000 B.C

α=	0.05
df	7	n-1

tc

2.364624

T.INV.2T(alpha,df)

CI = (X bar - tc*(S/SQRT(n)), X bar + tc*(S/SQRT(n)))

= (127.25 - 2.3646*(5.2847/SQRT(8)), 127.25 - 2.3646*(5.2847/SQRT(8)))

= (122.8319, 131.6681)

4000 B.C skull breadths: 122.83 to 131.67

95% CI for 150 A.D

α=	0.05
df	7	n-1

tc

2.364624

T.INV.2T(alpha,df)

CI = (X bar - tc*(S/SQRT(n)), X bar + tc*(S/SQRT(n)))

= (134.5 - 2.3646*(4.7809/SQRT(8)), 134.5 + 2.3646*(4.7809/SQRT(8)))

= (130.5031, 138.4969)

150 A.D skull breadths: 130.50 to 138.50