Question

Will fan and medal! Please help!
An inspector inspects large loads of potatoes to determine the proportion p in the shipment with major defects prior to making potato chips. She selects an SRS of 50 potatoes from over 2000 in the load. Only 2 of the potatoes sampled are found to have major defects.

Suppose you instead wish to test these hypotheses:
H0:mu=0.10
HAmu<0.10

What is the p-value for this test?
Choose one answer.
A. 0.9835
B. 0.1573
C. 0.0786
D. 0.9214
E. zero

Answer 1

@agent0smith

Answer 2

@jim_thompson5910

Answer 3

Are they giving you a table to find the p value?

Answer 4

No, that question is all the information I was given.

Answer 5

Do they expect you to use a table in the back of your book maybe?

Answer 6

I am taking the course online thus there is no back of the book.

Answer 7

hmm frustrating

Answer 8

they should have at least provided a table or a calculator maybe

Answer 9

anyways, do you know how to find the z-score?

Answer 10

Yes, but not with the information supplied.

Answer 11

You use the formula

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

In this case

p = 2/50 = 0.04
P = 0.10 (based on the null hypothesis)
n = 50

Answer 12

so z=-1.4142 ?

Answer 13

now use this calculator

http://www.danielsoper.com/statcalc3/calc.aspx?id=53

Answer 14

in this case,

mu = 0
sigma = 1

Answer 15

then you type -1.4142 into that last box

Answer 16

I got 0.07865159

Answer 17

me too

Answer 18

that's the approximate p value

Answer 19

basically the area under the curve to the left of -1.4142

Answer 20

Great, thanks!
Can you help with this one?
A psychic was tested for ESP. The psychic was presented with 400 cards face down and was asked to determine if the card was one of four symbols: a cross, a star, a circle, or a square. The psychic was correct in 120 of the cases. Let p represent the probability that the psychic correctly identifies the symbol on the card in a random trial.

How large a sample n would you need to estimate p with a margin of error of 0.01 with 95% confidence? (Use the guess 0.25 as the value of p)
Choose one answer.
A. N = 447
B. N = 7203
C. N = 9604
D. N = 30

Answer 21

check out this page

http://www.ltcconline.net/greenl/courses/201/estimation/ciprop.htm

Answer 22

what you need is at the very bottom, but it doesn't hurt to read the other stuff (since you may need it someday)

Answer 23

What do zsubc and E represent?

Answer 24

E is the margin of error

Answer 25

zsubc is the z critical value

Answer 26

in the case of the 95% confidence interval, it's roughly 1.96 since 95% of the distribution is between z = -1.96 and z = 1.96

Answer 27

So how do I get the critical value?

Answer 28

Well you either

* memorize it
* use a table (an inverse cdf table)
* use a calculator

Answer 29

here's a table you can use

http://3.bp.blogspot.com/_5u1UHojRiJk/TEdJJc6of2I/AAAAAAAAAIE/Ai0MW5VgIhg/s1600/t-table.jpg

Answer 30

notice in the row that starts with \(\Large \infty\), we have 1.960 right above the 95% confidence interval

Answer 31

So is 1.960 the critical value?

Answer 32

yes

Answer 33

So I went through it and got n=7203, is that correct?

Answer 34

I got the same

0.25*0.75*(1.96/0.01)^2 = 7,203