Question

In 1965, about 44% of the U.S. adult population had never smoked cigarettes. A national health survey of 1205 U.S. adults (presumably selected randomly) during 2006 revealed that 615 had never smoked cigarettes.
Reference: Ref 18-5

A 96% confidence interval for the proportion of U.S. adults in 2006 that have never smoked is
A. 0.461 to 0.560.
B. 0.470 to 0.551.
C. 0.481 to 0.540.
D. 0.487 to 0.534.

Answer 1

@jim_thompson5910

Answer 2

this sounds familiar...

Answer 3

Hmm lets see....

Answer 4

we did this one?

Answer 5

In 1965, about 44% of the U.S. adult population had never smoked cigarettes. A national health survey of 1205 U.S. adults (presumably selected randomly) during 2006 revealed that 615 had never smoked cigarettes.
Reference: Ref 18-5

Suppose you wished to test whether there has been a change since 1965 in the proportion of U.S. adults that have never smoked cigarettes. Which of the following are the appropriate hypotheses?
A. H0: p = 0.44, Ha: p > 0.44
B. H0: p = 0.51, Ha: p 0.51
C. H0: p = 0.44, Ha: p 0.44
D. H0: = 0.44, Ha: 0.44
Correct
Points Earned: 1/1
Your Response: C
4. In 1965, about 44% of the U.S. adult population had never smoked cigarettes. A national health survey of 1205 U.S. adults (presumably selected randomly) during 2006 revealed that 615 had never smoked cigarettes.
Reference: Ref 18-5

The P-value of the test of hypotheses in question 3 is
A. greater than 0.10.
B. between 0.05 and 0.10.
C. between .01 and 0.05.
D. below 0.01.

Answer 6

We did these :0

Answer 7

oh gotcha

Answer 8

good memory :D.

Answer 9

I'm going with B!

Answer 10

B

Answer 11

b

Answer 12

nope, not B

Answer 13

kay dont hate me

Answer 14

dang it! i thought i had it

Answer 15

sowwwyy

Answer 16

Hi team, thanks for the backup!

Answer 17

oh for sure!(;

Answer 18

you got it kboyyy

Answer 19

See you later in chat QT pies.

Answer 20

Ok back to this stupid stats.... We gotta do some P<Z stuff right?

Answer 21

or is this the confidence level BS...?

Answer 22

im never going back on chat EVER

Answer 23

me either!!!

Answer 24

NormalCfl(Wolfram).

Answer 25

we got kicked off... hahahahha

Answer 26

sooooo BA

Answer 27

You're cray... What did you say?

Answer 28

i told someone to die

Answer 29

sample proportion:

m = x/n = 615/1205 = 0.510373


Margin of error:

E = C*sqrt((m*(1-m))/n)

E = 2.0537489*sqrt((0.510373*(1-0.510373))/1205)

E = 2.0537489*sqrt((0.510373*(1-0.510373))/1205)

E = 0.02957538

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

96% Confidence interval:

(m - E, m + E)

(0.510373 - 0.02957538, 0.510373 + 0.02957538)

(0.48079762, 0.53994838)

(0.481, 0.540)

Answer 30

So It's c....

Answer 31

I'm going to write this formula down so I remember it...

Answer 32

Okay team onto the next Q!

Answer 33

okay!