Given the following information about glucose levels (in milligrams of glucose per 100 milliliters of blood.) ( from 9.5)

col1 73 61 104 75 85 65 62 98 92 106 ( non pregnant women)

col2 72 84 90 95 66 70 79 85 (pregnant women)

a. find the sample standard deviation s of the non pregnant women sample
b. find the sample standard deviation s of the pregnant women sample.
c. claim: " the variance of the glucose test results for non pregnant women is different to the variance for pregnant women"

notation: sigma squared is the population variance. (choose from 4 options below)

a. sigma squared of non pregnant women is greater than sigma squared of non pregnant women
b. sigma squared of non pregnant women is equal to sigma squared of non pregnant women
c. sigma squared of non pregnant women is not equal to sigma squared of non pregnant women
d. sigma squared of non pregnant women is less than sigma squared of non pregnant women

A. 16.385

B. 9.532

C. (16.358)^2 is different from (9.532)^2

a. sigma squared of non pregnant women is GREATER THAN sigma squared of pregnant women

Step-by-step explanation:

The formula for the standard deviation is:

s = √(∑(x - μ)²/(n))

where,

∑ = is the sum function

x = a number from the set

μ  = mean of the set

n =  is the amount of the numbers in the set.

the column of non-pregnant women is

[73, 61, 104, 75, 85, 65, 62, 98, 92, 106]

mean = μ  = (sum of all the numbers)/(amount of numbers)

μ = [73 + 61 + 104 +75 + 85 + 65 + 62 + 98 + 92 + 106]/(10)

μ = 821/10

μ = 82.1

similarly, for pregnant women the mean is

μ = 80.125

for non-pregnant you will get:

[ 82.81, 445.21, 479.61,  50.41,   8.41, 292.41, 404.01, 252.81,  98.01, 571.21]

for pregnant you will get:

[ 66.015,  15.015,  97.515, 221.265, 199.51,  102.515,   1.265,  23.76]

for non-pregnant = 2684.89/10

non-pregnant = 268.4 (this is the variance and it is denoted by s²)

for standard deviation 's', just take the square root of the variance

= 16.385.

similarly standard deviation of pregnant women can be found to be:

= 9.532.

A. non-pregnant 'S' =  16.385

B. pregnant 'S' = 9.532

you only have to show whether the variance of the above two columns are different or not.

And YES, the variances of the two column are indeed different, hence you make the CLAIM as written in question(C)

here you need to show how different are the two values:

recall the variances:

Column1 = 268.4 (non-pregnant)

Column2 = 9.532 (pregnant)

now you know that the variance of non-pregnant is GREATER THAN the variance of pregnant

the answer is:

the answer is:

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

then, calculating we have:

hence, the answer is:

have a nice day!

0

step-by-step explanation:

equals the constant of proportionality.

solve for