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KonradZuse

  • 2 years ago

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.

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  1. KonradZuse
    • 2 years ago
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    @jim_thompson5910

  2. jim_thompson5910
    • 2 years ago
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    this sounds familiar...

  3. KonradZuse
    • 2 years ago
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    Hmm lets see....

  4. jim_thompson5910
    • 2 years ago
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    we did this one?

  5. KonradZuse
    • 2 years ago
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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.

  6. KonradZuse
    • 2 years ago
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    We did these :0

  7. jim_thompson5910
    • 2 years ago
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    oh gotcha

  8. KonradZuse
    • 2 years ago
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    good memory :D.

  9. KonradZuse
    • 2 years ago
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    I'm going with B!

  10. itsssjennaaa
    • 2 years ago
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    B

  11. KG1234
    • 2 years ago
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    b

  12. jim_thompson5910
    • 2 years ago
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    nope, not B

  13. KG1234
    • 2 years ago
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    kay dont hate me

  14. itsssjennaaa
    • 2 years ago
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    dang it! i thought i had it

  15. itsssjennaaa
    • 2 years ago
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    sowwwyy

  16. KonradZuse
    • 2 years ago
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    Hi team, thanks for the backup!

  17. itsssjennaaa
    • 2 years ago
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    oh for sure!(;

  18. KG1234
    • 2 years ago
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    you got it kboyyy

  19. KonradZuse
    • 2 years ago
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    See you later in chat QT pies.

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

  21. KonradZuse
    • 2 years ago
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    or is this the confidence level BS...?

  22. KG1234
    • 2 years ago
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    im never going back on chat EVER

  23. itsssjennaaa
    • 2 years ago
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    me either!!!

  24. KonradZuse
    • 2 years ago
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    NormalCfl(Wolfram).

  25. itsssjennaaa
    • 2 years ago
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    we got kicked off... hahahahha

  26. KG1234
    • 2 years ago
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    sooooo BA

  27. KonradZuse
    • 2 years ago
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    You're cray... What did you say?

  28. KG1234
    • 2 years ago
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    i told someone to die

  29. jim_thompson5910
    • 2 years ago
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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)

  30. KonradZuse
    • 2 years ago
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    So It's c....

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

  32. KonradZuse
    • 2 years ago
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    Okay team onto the next Q!

  33. itsssjennaaa
    • 2 years ago
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    okay!

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