Question

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(1 point) Archeologists have discovered a sample of ancient Egyptian skulls from 150 A.C. A sample of 10 Egyptian skull breadth (in mm) from 150 A.C. is show below: Skull breadths Sample: 135, 127, 138, 127, 128, 141, 130, 128, 132, 134 - Part 1: 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: 132 150 A.C. skull breadths standard deviation: 4.8990 - Part 2 Using your answers from part 1 and the formula for a 98% 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 x where x is any number, type sqrt(x). For example, V2 should be written as sqrt(2). 150 A.C. skull breadths: to
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Answer 1

Solution-:

(1) Part -1 By using R-Software we find mean and SD:

> x=c(135,127,138,127,128,141,130,128,132,134);x
[1] 135 127 138 127 128 141 130 128 132 134
> #For mean
> Mean=mean(x);Mean
[1] 132
> #for Standard Deviation
> sd=sd(x);sd
[1] 4.898979

From this output we get,

Skull breadth mean$\bar{x}=132$ and  Skull breadth standard deviation $s=4.8990$

(1) Part -2

The 98 % C.I. is gien by;

here, n=10

$(\bar{x}- t_{(n-1),\alpha/2}*\frac{s}{\sqrt{n}},\bar{x}+ t_{(n-1),\alpha/2}*\frac{s}{\sqrt{n}})$

$(132- 2.821*\frac{4.8990}{\sqrt{10}},132+ 2.821*\frac{4.8990}{\sqrt{10}})$

$t_{(n-1),\alpha/2}=t_{(10-1),0.02/2}=t_{9,0.01}=2.821$

By using MS-Excel Function "=TINV(0.02,9)")

The required C.I. is

$(132- 2.821*\frac{4.8990}{\sqrt{10}},132+ 2.821*\frac{4.8990}{\sqrt{10}})$

$(132- 2.821*\frac{4.8990}{sqrt{(10)}},132+ 2.821*\frac{4.8990}{sqrt{(10)}})$

(2) Part -1 By using R-Software we find mean and SD:

> y=c(136,124,137,122,120,119);y
[1] 136 124 137 122 120 119
> #For mean
> Mean=mean(y);Mean
[1] 126.3333
> #for Standard Deviation
> sd=sd(y);sd
[1] 8.066391

From this output we get,

Skull breadth mean$\bar{x}=126.33$ and  Skull breadth standard deviation $s=8.0664$

(2) Part -2

The 99 % C.I. is gien by;

here, n=6

$(\bar{x}- t_{(n-1),\alpha/2}*\frac{s}{\sqrt{n}},\bar{x}+ t_{(n-1),\alpha/2}*\frac{s}{\sqrt{n}})$

$(126.33- 4.032*\frac{8.0664}{\sqrt{6}},126.33+ 4.032*\frac{8.0664}{\sqrt{6}})$

$t_{(n-1),\alpha/2}=t_{(6-1),0.01/2}=t_{5,0.005}=4.032$

By using MS-Excel Function "=TINV(0.01,5)")

The required C.I. is

$(126.33- 4.032*\frac{8.0664}{\sqrt{6}},126.33+ 4.032*\frac{8.0664}{\sqrt{6}})$

$(126.33- 4.032*\frac{8.0664}{sqrt{(6)}},126.33+ 4.032*\frac{8.0664}{sqrt{(6)}})$