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.

- KonradZuse

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- schrodinger

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- KonradZuse

@jim_thompson5910

- jim_thompson5910

this sounds familiar...

- KonradZuse

Hmm lets see....

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## More answers

- jim_thompson5910

we did this one?

- KonradZuse

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.

- KonradZuse

We did these :0

- jim_thompson5910

oh gotcha

- KonradZuse

good memory :D.

- KonradZuse

I'm going with B!

- anonymous

B

- anonymous

b

- jim_thompson5910

nope, not B

- anonymous

kay dont hate me

- anonymous

dang it! i thought i had it

- anonymous

sowwwyy

- KonradZuse

Hi team, thanks for the backup!

- anonymous

oh for sure!(;

- anonymous

you got it kboyyy

- KonradZuse

See you later in chat QT pies.

- KonradZuse

Ok back to this stupid stats.... We gotta do some P

- KonradZuse

or is this the confidence level BS...?

- anonymous

im never going back on chat EVER

- anonymous

me either!!!

- KonradZuse

NormalCfl(Wolfram).

- anonymous

we got kicked off... hahahahha

- anonymous

sooooo BA

- KonradZuse

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

- anonymous

i told someone to die

- jim_thompson5910

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)

- KonradZuse

So It's c....

- KonradZuse

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

- KonradZuse

Okay team onto the next Q!

- anonymous

okay!

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