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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.

  • one year ago
  • one year ago

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  1. KonradZuse
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    @jim_thompson5910

    • one year ago
  2. jim_thompson5910
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    this sounds familiar...

    • one year ago
  3. KonradZuse
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    Hmm lets see....

    • one year ago
  4. jim_thompson5910
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    we did this one?

    • one year ago
  5. KonradZuse
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    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.

    • one year ago
  6. KonradZuse
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    We did these :0

    • one year ago
  7. jim_thompson5910
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    oh gotcha

    • one year ago
  8. KonradZuse
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    good memory :D.

    • one year ago
  9. KonradZuse
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    I'm going with B!

    • one year ago
  10. itsssjennaaa
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    B

    • one year ago
  11. KG1234
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    b

    • one year ago
  12. jim_thompson5910
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    nope, not B

    • one year ago
  13. KG1234
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    kay dont hate me

    • one year ago
  14. itsssjennaaa
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    dang it! i thought i had it

    • one year ago
  15. itsssjennaaa
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    sowwwyy

    • one year ago
  16. KonradZuse
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    Hi team, thanks for the backup!

    • one year ago
  17. itsssjennaaa
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    oh for sure!(;

    • one year ago
  18. KG1234
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    you got it kboyyy

    • one year ago
  19. KonradZuse
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    See you later in chat QT pies.

    • one year ago
  20. KonradZuse
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    Ok back to this stupid stats.... We gotta do some P<Z stuff right?

    • one year ago
  21. KonradZuse
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    or is this the confidence level BS...?

    • one year ago
  22. KG1234
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    im never going back on chat EVER

    • one year ago
  23. itsssjennaaa
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    me either!!!

    • one year ago
  24. KonradZuse
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    NormalCfl(Wolfram).

    • one year ago
  25. itsssjennaaa
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    we got kicked off... hahahahha

    • one year ago
  26. KG1234
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    sooooo BA

    • one year ago
  27. KonradZuse
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    You're cray... What did you say?

    • one year ago
  28. KG1234
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    i told someone to die

    • one year ago
  29. jim_thompson5910
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    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)

    • one year ago
  30. KonradZuse
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    So It's c....

    • one year ago
  31. KonradZuse
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    I'm going to write this formula down so I remember it...

    • one year ago
  32. KonradZuse
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    Okay team onto the next Q!

    • one year ago
  33. itsssjennaaa
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    okay!

    • one year ago
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