Question

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Answer 1

Group n x s
A 150 $1987 $392
B 150 $2056 $413


a) Do the data show a significant difference between the mean amounts charged by customers offered the two plans? Carry out a complete test.
b) The distributions of amounts charged are skewed to the right, but outliers are prevented by the limits that the bank imposes on credit balances. Do you think that skewness threatens the validity of the test? Explain your answer.

Answer 2

@jim_thompson5910

Answer 3

let me think for a sec

Answer 4

ok

Answer 5

ok to start you off, read this page on two sample t tests

http://ccnmtl.columbia.edu/projects/qmss/the_ttest/twosample_ttest.html

this will refresh your memory on the whole process (which hopefully you have learned it before)

Answer 6

Okk

Answer 7

I did

Answer 8

ok and it looks familiar?

Answer 9

ya

Answer 10

ok luckily there's a calculator on that page as well

Answer 11

ok

Answer 12

see where it says

Two-Sample
T-Test Calculator

Answer 13

click that and enter the data

tell me what p value you get

Answer 14

ok i put in the data and got results.. t = 1.4841
df = 298
standard error of difference = 46.493

Answer 15

@jim_thompson5910

Answer 16

one sec while i check your results

Answer 17

read me the p value please

it's near where it says "P value and statistical significance: "

Answer 18

its .1388?

Answer 19

good

Answer 20

ok cool. what do i put for a

Answer 21

so at the 5% significance level (the default level), it's NOT significant (the p value needs to be less than 0.05)

even at the 10% significance level, it's still NOT significant (still not smaller than 0.10)

so the data is NOT statistically significant....which means that the two sample means aren't statistically significantly different

Answer 22

Ok yea! so for a, id put all that?^

Answer 23

Do the data show a significant difference between the mean amounts charged by customers offered the two plans?

no there's no significant difference between the two sample means

Answer 24

that's basically what you would say more or less

of course you would throw in the test (ie the steps needed to get there)

Answer 25

Oky cool thx! wat about B

Answer 26

b?

Answer 27

it's hard to say, but even though there's a right skew, it's not that bad because of the imposed limits

Answer 28

so I would think that the skew doesn't alter the fact that these distributions are approximately normal (more or less)

Answer 29

thats what i was thinking.. ok thanks! i have another one:

Answer 30

this is also a two sample t test

Answer 31

so follow the same basic procedure

1) use the calculator given above

2) read off the p value to make the proper decisions

Answer 32

okay umm

Answer 33

if it helps, think of the two groups as A and B

Answer 34

The two-tailed P value equals 0.0172
?

Answer 35

very good, getting the same

Answer 36

so...

is 0.0172 smaller than 0.05 ?

is 0.0172 smaller than 0.01 ?

Answer 37

Yes and no for the second

Answer 38

perfect

Answer 39

Do beta-blockers reduce the pulse rate at the 5% level? At the 1% level?



yes for the 5% level (since it's statistically significant at the 5% level)

but no for the 1% (its not statistically significant at the 1% level)

Answer 40

see how I'm getting all this?

Answer 41

Okay ya

Answer 42

b?

Answer 43

ok sadly this calc only does 95% confidence intervals

Answer 44

so one sec

Answer 45

oh ok

Answer 46

does it say anywhere that the population variances are assumed to be equal?

Answer 47

Nope i dont see it

Answer 48

ok i'll just guess "no, they aren't assumed to be equal"

Answer 49

Ok

Answer 50

ok first thing we need to do is calculate the standard error

which we'll call SE

Answer 51

Ok

Answer 52

SE = sqrt( ((s1)^2)/(n1) + ((s2)^2)/(n2) )


SE = sqrt( ((7.8)^2)/(30) + ((8.3)^2)/(30) )


SE = 2.07950314578587


SE = 2.0795

Answer 53

ok got that

Answer 54

then we need to find the critical value

we would use a calculator to do this

http://www.wolframalpha.com/input/?i=inverse+t&a=*MC.inverse+t-_*Formula.dflt-&f2=29&f=StudentTProbabilities.df_29&f3=0.005&x=7&y=9&f=StudentTProbabilities.pr_0.005&a=*FVarOpt.1-_***StudentTProbabilities.pr--.***StudentTProbabilities.x--.**StudentTProbabilities.l-.*StudentTProbabilities.r---.*--

notice on the second line it says "2.756"

so that's our critical value

Answer 55

ok

Answer 56

now onto the actual confidence interval


Lower Limit:

L = (xbar1 - xbar2) - t*SE

L = (65.2-70.3) - 2.756*(2.0795)

L = -10.831102


Upper Limit:


U = (xbar1 - xbar2) + t*SE

U = (65.2-70.3) + 2.756*(2.0795)

U = 0.631102

Answer 57

So the confidence interval

(L, U)

turns into

(-10.831102, 0.631102)

Answer 58

which means the 99% CI is (-10.831102, 0.631102)


CI = confidence interval

Answer 59

ok!

Answer 60

ok i have something else:

Answer 61

@jim_thompson5910

Answer 62

?

Answer 63

sry one sec

Answer 64

still looking up formula

Answer 65

ok

Answer 66

hmm does it say anywhere that you have to assume that p-hat = 0.5 ?

Answer 67

oh wait, p = 0.25 nvm

Answer 68

ok

Answer 69

so we would do this...


n = p(1-p)(z/E)^2


n = 0.25(1-0.25)(z/E)^2


n = 0.25(0.75)(z/E)^2


n = 0.1875(z/E)^2


n = 0.1875(1.96/0.05)^2


n = 0.1875(39.2)^2


n = 0.1875(1536.64)


n = 288.12


n = 289 ... Always round UP!!! (to the nearest whole number)

Answer 70

Ok!

Answer 71

is that it?

Answer 72

so the min sample size is 289

Answer 73

yep

Answer 74

ok one sec!

Answer 75

same idea, but different numbers

Answer 76

oh wait, they want the margin of error this time

Answer 77

well it's a similar idea at least

Answer 78

okk

Answer 79

soo

Answer 80

actually this is a two part problem

the first part asks for the min sample size

Answer 81

the second part asks for the margin of error

Answer 82

Ok

Answer 83

so for the first part, we use the formula

n = p(1-p)(z/E)^2

where

p = 0.50
z = 1.96
E = 0.08

Answer 84

ok

Answer 85

n = .50(1-.50)(1.96/.08)^2

Answer 86

actually my bad,

p = 0.60
z = 1.96
E = 0.08

sorry about that

Answer 87

ah ok lol hold on

Answer 88

144.06??

Answer 89

@jim_thompson5910

Answer 90

so n = 145

Answer 91

that's the min sample size needed

Answer 92

now we use this to answer the second part of the question

Answer 93

ok cool

Answer 94

using this formula

E = z*sqrt( (p*(1-p))/n )

in this case

z = 1.96 (still the same since we're still dealing with a 95% CI)
p = 0.50 (this changes because they tell us it does)
n = 145 (we just found this)

Answer 95

ok

Answer 96

what do you get

Answer 97

is it 0.08138457025