Question

see attachment.

Answer 1

@jim_thompson5910 Last Question!

Answer 2

do you know the confidence interval formulas?

Answer 3

is it this?

Answer 4

correct

Answer 5

ok on sec i'll try that.

Answer 6

ok

Answer 7

ok i'm stuck. were do i get t from?

Answer 8

t is the critical value

Answer 9

this is the value k that satisfies the equation

P(T > k) = 0.025

where df = n-1 = 34 - 1 = 33


Note: alpha = 1-CL = 1-0.95 = 0.05, so alpha/2 = 0.05/2 = 0.025

Answer 10

i can't find a big enough table with 33 on it.

Answer 11

luckily we have a calculator

http://www.wolframalpha.com/input/?i=tcdf&a=*C.tcdf-_*Formula.dflt-&f2=33&x=0&y=0&f=StudentTProbabilities.df_33&f3=0.025&f=StudentTProbabilities.pr_0.025&a=*FVarOpt.1-_***StudentTProbabilities.pr--.***StudentTProbabilities.x--.**StudentTProbabilities.l-.*StudentTProbabilities.r---.*--

Answer 12

look on line 2 under the "x values" section

Answer 13

and you'll see the value of 2.035, this is the critical value

Answer 14

so now i do 17.6+2.035*4.4/sq(35)?

Answer 15

n = 34, not n = 35

Answer 16

so the upper limit is 17.6+2.035*4.4/sq(34)


the lower limit is 17.6-2.035*4.4/sq(34)

Answer 17

you compute them out and evaluate as much as possible of course

Answer 18

17.6+2.035*4.4/sq(34)=19.13559833
17.6-2.035*4.4/sq(34)= 16.06440167

Answer 19

then round each to 2 decimal places

Answer 20

so 16.06,19.14?

Answer 21

yep

Answer 22

for part b, the same steps will be taken but this time n = 51 so the critical value will change and it will become t = 0.962


http://www.wolframalpha.com/input/?i=tcdf&a=*C.tcdf-_*Formula.dflt-&f2=50&x=5&y=8&f=StudentTProbabilities.df_50&f3=0.025&f=StudentTProbabilities.pr \u005f0.025&a=*FVarOpt.1-_***StudentTProbabilities.pr--.***StudentTProbabilities.x--.**StudentTProbabilities.l-.*StudentTProbabilities.r---.*--

Answer 23

so which one do i look at?

Answer 24

nvm, the calculator is giving me trouble

one sec while I find another one

Answer 25

hows this?

Answer 26

yeah i was getting the same thing, but the probability kept changing on me

Answer 27

I'm using this one now

http://www.danielsoper.com/statcalc3/calc.aspx?id=10

and got a critical value of t = 2.00855911

so it's roughly t = 2.0086

Answer 28

so i do 17.6+2.0086*4.4/sq(50)
& 17.6-2.0086*4.4/sq(50)?

Answer 29

n = 51

the df =50

Answer 30

so you're mixing the two up a bit

Answer 31

oh ok i see that now.

Answer 32

16.36,18.84?

Answer 33

I'm getting 16.35 instead of 16.36

Answer 34

oh wait, using the wrong value, one sec

Answer 35

ok now I'm getting 16.36 and 18.84

Answer 36

so you are correct

Answer 37

ok so now i'm not sure how the margin of error is affected.

Answer 38

well the margin of error is basically how far you go from the center of the confidence interval to either endpoint

Answer 39

so what is the margin of error for the CI in part a

Answer 40

i have no idea

Answer 41

what is the midpoint of the CI in part a

Answer 42

lol i don't know this is confusing. your the genies here give me a hint. lol

Answer 43

If E is the margin of error, then

Lower Limit = xbar - E

Upper Limit = xbar + E


brb

Answer 44

so i do 16.36=17.6-E
&18.84=17.6+E?

Answer 45

solve each equation for E to get

16.36=17.6-E

16.36-17.6=17.6-E-17.6

-1.24 = -E

E = 1.24

------------------------

18.84=17.6+E

18.84-17.6=E

1.24=E

E = 1.24

So the margin of error for the confidence interval in part b) is 1.24

Answer 46

the margin of error for the CI in part a) is

U = xbar + E

19.14 = 17.6 + E

E = 19.14 - 17.6

E = 1.54

-----------------

L = xbar - E

16.06 = 17.6 + E

-E = 16.06 - 17.6

-E = -1.54

E = 1.54

So the margin of of error for the confidence interval in part a) is 1.54

Answer 47

this is what you expect

larger sample size ----> smaller margin of error

Answer 48

So the margin of error decreases?

Answer 49

yep

Answer 50

as your sample gets bigger, your get a more accurate picture of what you're trying to estimate

Answer 51

ok now for part c

Answer 52

for part c), the critical value is t = 2.7333

Answer 53

15.54,19.66?

Answer 54

good

Answer 55

now find the margin of error and compare it to the margin of error in part a

Answer 56

E=-2.06
E=2.06

Answer 57

so your margin of error is 2.06, compare that to the margin of error for part a

Answer 58

it increased

Answer 59

so for part c, the answer is A

Answer 60

ok Now for the last part. THANK GOD!! lol it's already 10pm!

Answer 61

part d is a bit odd...hmm let me think

Answer 62

I'm assuming it's a 99% CI, but it doesn't say

Answer 63

I'm thinking it's B because you can still use small sample sizes (I think) to compute the CI

Answer 64

I think the only thing you need to worry about is if the data is normally distributed

Answer 65

ok Thanks for the help!!

Answer 66

yw