Question

A researcher used a new drug to treat 100 subjects with high cholesterol. For the patients in the study, after two months of treatment the average decrease in cholesterol level was 80 milligrams per deciliter (mg/dl). Assume that the decrease in cholesterol after two months of taking the drug follows a Normal distribution, with unknown mean μ and standard deviation σ = 20 mg/dl. The researcher will construct a 90% confidence interval to estimate μ.

Answer 1

The researcher's 90% confidence interval for μ is
A. 78 mg/dl to 82 mg/dl.
B. 76.71 mg/dl to 83.29 mg/dl.
C. 60 mg/dl to 100 mg/dl.
D. 47.1 mg/dl to 112.9 mg/dl.

The margin of error associated with a 90% confidence interval for the researcher's 90% confidence interval for μ is
A. 1.645 mg/dl.
B. 20 mg/dl.
C. 2 mg/dl.
D. 3.29 mg/dl.

Answer 2

@jim_thompson5910

Answer 3

I'm thinkinh A for the first part.

Answer 4

this problem is a bit backwards because it helps to find the margin of error FIRST before you try to find the confidence interval

Answer 5

Does stats ever not be backwards? :)

Answer 6

true

Answer 7

the margin of error is

E = k*sigma/sqrt(n)

where k is the value such that P(-zk < Z < k) = 0.90

Answer 8

oops meant to say P(-k < Z < k) = 0.90

Answer 9

>(

Answer 10

E = k*sigma/sqrt(n)

E = 1.645*20/sqrt(100) ... see note

E = 3.29

Note: see this page: http://www.wolframalpha.com/input/?i=invnorm%280.90%29

Answer 11

I was trying to find a hw question that was similar, I forgot the formula it says.

Answer 12

Wolfram's the best :)

Answer 13

it is

you then use E to find the confidence interval

lower bound: xbar - E

upper bound: xbar + E

Answer 14

shouldn't we put a mean and SD in or not needed?

Answer 15

oh... unknown mean woops.

Answer 16

i put in sigma, which is the SD

Answer 17

the population mean is unknown, but we can use xbar

Answer 18

wolfram says 1, this says .2?

Answer 19

x values are 1.2 it looks like...

Answer 20

huh?

Answer 21

sorry :P

Answer 22

yeah i'm completely lost lol

Answer 23

I'm more lost LOL.

Answer 24

well do you see how to get E?

Answer 25

yessir.

Answer 26

now xbar is which?

Answer 27

1.282?

Answer 28

where are you getting that value?

Answer 29

Wolfram :P

Answer 30

give me the link

Answer 31

of what you're seeing

Answer 32

the one you posted :p.

Answer 33

where it says "x-values."

Answer 34

oh you're picking the wrong value

Answer 35

you want P( |z| < 1.645)

Answer 36

yeah I was just thinking that.

Answer 37

oic...

Answer 38

conveniently named "confidence level"

Answer 39

LOL woops.

Answer 40

np

Answer 41

3.29 - 1.645 and 3.29 + 1.645?

Answer 42

no

Answer 43

1.645 isn't the margin of error

Answer 44

it helps you find it though

Answer 45

>(

Answer 46

so that's not xbar ok... :(

Answer 47

E = k*sigma/sqrt(n)

E = 1.645*20/sqrt(100)

E = 3.29

Answer 48

xbar is given to be 80

Answer 49

ok......

Answer 50

confidence interval

lower bound: xbar - E

upper bound: xbar + E

Answer 51

okay that's what I was thinkign from the start, this stats always konfuses me... :(.

Answer 52

So B?

Answer 53

you'll get it, just practice more

Answer 54

B and D, yes

Answer 55

oic E is the answer to the second part....

Answer 56

yeah exactly, which is why it's confusing to have them in this order

Answer 57

oic and k is our confidence level.....

Answer 58

sorta

Answer 59

>(

Answer 60

k is determined by the confidence level

Answer 61

but k = 1.645 isn't the confidence level itself

Answer 62

because we put .9?

Answer 63

that's the confidence level

Answer 64

90%

Answer 65

well that's what you posted before k = 1.645?

Answer 66

ah okay....

Answer 67

yeah but didn't say/mean it was the confidence level, it's just based off it

Answer 68

CL changes -----> k changes

Answer 69

yeah.... It just says confidence level in Wolfram :P.....

Answer 70

true it does say that, just gotta know which value they're referring to

Answer 71

*throws laptop*