Question

how do you find the p value in statistics

Answer 1

isnt" p" the polulation, or portion or sample size

Answer 2

can you help me solve this ...

Answer 3

The alternative hypothesis for the difference between two means is Ha:µ1-µ2 ≠ 0. For this test, the test statistic, z is 2.53. What is the p-value for the test?

Answer 4

p is the confidence level at which the test value equals the critical value

Answer 5

ok how can i apply this to the question i posted above?

Answer 6

difference between 2 means is a 2 tailed test

Answer 7

find the area under of z between -2.53 and 2.53

Answer 8

under the standard normal curve for -2.53 < z < 2.53

Answer 9

im sorry, how do we find the area under z?

Answer 10

from a standard normal table

Answer 11

http://statistics.laerd.com/statistical-guides/img/normal/normal-table-large.png

Answer 12

alright thanks, i have no idea how any of this works. im doin the work for a friend.

Answer 13

the area to the right of z=2.53 is (1-.9943) = .0057
the p value is twice that, p = .0114

Answer 14

wow thanks so much!!! i really appreciate it. would you be able to help me with one more question?

Answer 15

sure

Answer 16

A study by a corporation revealed that the mean weekly salary for managers in department A was $1,500 with a standard deviation of $100 (sample size = 64) while the mean weekly salary for managers in department B was $1,275 and a standard deviation of $150 ( sample size = 81 ). Construct the 95% confidence interval for the difference between the mean salaries for managers.

Answer 17

i even have four choices for this one.. ill send them right now

Answer 18

(167.3, 234.2 )
(202.3, 227.1)
(104.9, 167.3 )
(184.2, 265.8)

Answer 19

alpha = 1 - .95 = .025

Answer 20

df = 64 - 1

Answer 21

t (alpha/2) = .679

Answer 22

from the table http://3.bp.blogspot.com/_C6375WoyYP0/THVqQkJRqbI/AAAAAAAAASU/ML7g-OPPgug/s1600/Student-t-table.png

Answer 23

actually because it's multiple choice, you can just pick the answer which has an average of (1500 - 1275) = 225. That's answer D

Answer 24

(184.2, 265.8) ? that's what we were thinking! wow thanks

Answer 25

i also read the table wrong

Answer 26

t (alpha/2) = 2.000

Answer 27

(1500 - 1275) + 2*Sqrt[100^2/64 + 150^2/81] and (1500 - 1275) - 2*Sqrt[100^2/64 + 150^2/81]
which is close to answer D

Answer 28

ok thanks so much.. thats what i was going with.. there is one more with the 95% interval.. would you mind helping again? if not its ok we appreciate it

Answer 29

Suppose you have a sample selected from a normally distributed population and the sample mean is 64.5 and the sample standard deviation is s = 7.6. Find a 95% confidence interval for µ assuming n=9.

Answer 30

there are also choices for this.... (53.78, 60.73)
(58.66,70.34)
(61.47, 67.21)
(62.16, 72.57)

Answer 31

alpha = 1 - .95 = .05 again

Answer 32

look up t from table with alpha/2 = .025 and df = 9 -1 = 8

Answer 33

t (alpha/2) = 2.306

Answer 34

the E = 2.306*7.6/Sqrt[9] = 5.84187

Answer 35

64.5+5.84187=70.3419
64.5-5.84187=58.6581

answer B

Answer 36

thank youu sooo muchhh!!

Answer 37

i got those charts from elementary statistics 9ed by mario triola

Answer 38

ok great gonna save them to my computer so that i have it for the future

Answer 39

you want the whole book?

Answer 40

no you helped us enough. i really appreciate it.. i actually have one more question- would you be willing to help.. its my last one

Answer 41

The mean fill of containers of sugar is 16.0 ounces with a standard deviation of 1.5 ounces. Samples of size n = 9 are taken from the population of all containers of sugar. The population values are normally distributed. Find the probability that a given sample mean will exceed 17 ounces. (Hint: See Example 6.7b in Section 6.3 of the text. Don't worry that the value of n is only 9.)

Approximately 0.09
Approximately 0.05
Less than 0.02
.04

Answer 42

working on it

Answer 43

the standard deviation for the sample mean is sigma/Sqrt[n] = 1.5/Sqrt[9] = 0.5

Answer 44

the mean of the sample = 16

Answer 45

calculate z-score (17-16)/.5 = 2

Answer 46

find the area to the right of z=2 on standard normal table

Answer 47

= (1-.9772) = .0228

Answer 48

doesn't seem to match the answer choices

Answer 49

hm i wonder why. thats strange!

Answer 50

I'd put C and ask the teacher

Answer 51

ok thank you for all your help today!