KonradZuse
  • 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 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.
Statistics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
KonradZuse
  • KonradZuse
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
this sounds familiar...
KonradZuse
  • KonradZuse
Hmm lets see....

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More answers

jim_thompson5910
  • jim_thompson5910
we did this one?
KonradZuse
  • 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
  • KonradZuse
We did these :0
jim_thompson5910
  • jim_thompson5910
oh gotcha
KonradZuse
  • KonradZuse
good memory :D.
KonradZuse
  • KonradZuse
I'm going with B!
anonymous
  • anonymous
B
anonymous
  • anonymous
b
jim_thompson5910
  • jim_thompson5910
nope, not B
anonymous
  • anonymous
kay dont hate me
anonymous
  • anonymous
dang it! i thought i had it
anonymous
  • anonymous
sowwwyy
KonradZuse
  • KonradZuse
Hi team, thanks for the backup!
anonymous
  • anonymous
oh for sure!(;
anonymous
  • anonymous
you got it kboyyy
KonradZuse
  • KonradZuse
See you later in chat QT pies.
KonradZuse
  • KonradZuse
Ok back to this stupid stats.... We gotta do some P
KonradZuse
  • KonradZuse
or is this the confidence level BS...?
anonymous
  • anonymous
im never going back on chat EVER
anonymous
  • anonymous
me either!!!
KonradZuse
  • KonradZuse
NormalCfl(Wolfram).
anonymous
  • anonymous
we got kicked off... hahahahha
anonymous
  • anonymous
sooooo BA
KonradZuse
  • KonradZuse
You're cray... What did you say?
anonymous
  • anonymous
i told someone to die
jim_thompson5910
  • 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
  • KonradZuse
So It's c....
KonradZuse
  • KonradZuse
I'm going to write this formula down so I remember it...
KonradZuse
  • KonradZuse
Okay team onto the next Q!
anonymous
  • anonymous
okay!

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