Question

The one-sample t statistic for a test of
H0: mean= 20
Ha: mean <20
based on n = 12 observations has the value t = -2.45.

Answer 1

(a) What are the degrees of freedom for this statistic?
(b) What is the exact p-value?

Answer 2

@jim_thompson5910

Answer 3

@hartnn @Mertsj

Answer 4

when doing a one sample t test, the degrees of freedom (df) are

df = n-1

Answer 5

the p value is the probability of observing the test statistic or any other extreme test statistic (assuming the null hypothesis is true)

Answer 6

ok

Answer 7

12-1?

Answer 8

df = n - 1

df = 12 - 1

df = 11

yep

Answer 9

degrees of freedom is 11

Answer 10

okk what about b?

Answer 11

you need to calculate the area under the curve to the left of t = -2.45 (df = 11)

Answer 12

how do i calculate it?

Answer 13

http://www.wolframalpha.com/input/?i=tcdf&a=*C.tcdf-_*Formula.dflt-&f2=11&x=0&y=0&f=StudentTProbabilities.df_11&f3=-2.45&f=StudentTProbabilities.x_-2.45&a=*FVarOpt.1-_***StudentTProbabilities.x--.***StudentTProbabilities.pr--.**StudentTProbabilities.l-.*StudentTProbabilities.r---.*--

Answer 14

using a calc of course lol, but the best choice is using wolfram alpha

Answer 15

ok what do i type in

Answer 16

? @jim_thompson5910

Answer 17

I typed it in for you and posted the link

Answer 18

ok what do i look at tho haha

Answer 19

but you would type in tcdf and then specify the endpoint and df

Answer 20

endpoint is -2.45

Answer 21

lol you would look at the probabilities section and the portion that has t < -2.45

Answer 22

df is 11

Answer 23

read me that number

Answer 24

i know

Answer 25

.01612

Answer 26

.01612
?

Answer 27

perfect, that's your p value

Answer 28

thanks so much ! i have one more prob ill typ ehere plz

Answer 29

ok

Answer 30

The amount of lead in a certain type of soil, when released by a standard extraction method, averages 86 parts per million (ppm). A new extraction method is tried on 40 specimens of the soil, yielding a mean of 83 ppm lead and a standard deviation of 10 ppm.

(a) Is there significant evidence at the 5% level that the new method frees less lead from the soil? What about the 1% level?
(b) A critic argues that because of variations in the soil, the effectiveness of the new method is confounded with characteristics of the particular soil specimens used. Briefly describe a better data production design that avoids this criticism.

Answer 31

what's the population mean mu

Answer 32

how do i find that?

Answer 33

?

Answer 34

have you done hypothesis tests before?

Answer 35

kinda..

Answer 36

?

Answer 37

what do i do for a

Answer 38

one sec

Answer 39

when it gives you

The amount of lead in a certain type of soil, when released by a standard extraction method, averages 86 parts per million (ppm). A new extraction method is tried on 40 specimens of the soil, yielding a mean of 83 ppm lead and a standard deviation of 10 ppm.

we can pull out these pieces of info


mu = 86
xbar = 83
s = 10
n = 40

Answer 40

ok cool

Answer 41

so are you able to use this info to conduct a hypothesis test

Answer 42

95 percent Confidence Interval: 83 plus/minus 2.021 * 10 / SQRT(10)
= (76.61, 89.39)

(a) 1 and 5 percent level: No! The new method does not free less lead from the soil.

(b)

Answer 43

is that right? if not can u tell me what to do

Answer 44

you're not doing confidence intervals here

Answer 45

you're doing a hypothesis test

Answer 46

so you need to find a p value (that's one way to do it)

Answer 47

ok! how do i do that with the info you gave above?

Answer 48

@jim_thompson5910 ?

Answer 49

well you would set up a normal distribution with mean 86 and standard deviation of 10/sqrt(40) = 1.5811

Answer 50

10/sqrt(40) = 1.5811 ok

Answer 51

then you would use this distribution to find the area under the curve to the left of 83

Answer 52

how do i do that? sorry lol ^ @jim_thompson5910

Answer 53

like how do i find a

Answer 54

use that calculator (the wolfram alpha calculator) we've been using

Answer 55

ok im on it

Answer 56

but this time we'll be using the normalcdf function

Answer 57

not the tcdf function

Answer 58

how do i put it at cdf

Answer 59

@jim_thompson5910

Answer 60

you type in normalcdf to get to the proper function you want

Answer 61

then you fiddle with the settings a bit to include the mean and standard deviation

I'll brb

Answer 62

ok i have to go soon but

Answer 63

whats a? @jim_thompson5910

Answer 64

@jim_thompson5910 r u back?????

Answer 65

yeah sry took longer than i thought

Answer 66

its ok. :P what is A?

Answer 67

the p value is given here

http://www.wolframalpha.com/input/?i=normalcdf&a=*C.normalcdf-_*Formula.dflt-&f2=83&x=1&y=9&f=NormalProbabilities.z_83&f3=86&f=NormalProbabilities.mu_86&f4=1.5811&f=NormalProbabilities.sigma_1.5811&a=*FVarOpt.1-_***NormalProbabilities.z--.***NormalProbabilities.pr--.**NormalProbabilities.l-.*NormalProbabilities.r---.**NormalProbabilities.mu-.*NormalProbabilities.sigma---

Answer 68

look at the line with z < 83

Answer 69

see it?

Answer 70

ya

Answer 71

.02889 @jim_thompson5910 ?

Answer 72

good

Answer 73

now ask yourself, is that larger or smaller than 0.05

Answer 74

smaller

Answer 75

so this means that it's significant at the 5% level

is .02889 larger or smaller than 0.01

Answer 76

larger

Answer 77

so we can say

it's significant at the 5% level

but it's not significant at the 1% level

Answer 78

ok got that!

Answer 79

ie we can reject the null at 5% significance but we cannot reject the null at 1% significance

Answer 80

oko so thats a, thans! what about b?

Answer 81

(b) A critic argues that because of variations in the soil, the effectiveness of the new method is confounded with characteristics of the particular soil specimens used. Briefly describe a better data production design that avoids this criticism.


To design a better experiment, the idea is to control all the variables the best you can. So try to get each sample to have the same characteristics so the affect of confounding variables is low. This means each sample gets the same amount of water, air, etc and is found in the same general location. This should lead to more consistent samples and a more accurate experiment.

Answer 82

Basically what you want to do is isolate one variable and test it while holding other variables fixed. This is so you know that they don't have an affect on the outcome of the experiment

Answer 83

thanks so much

Answer 84

np