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.

@jim_thompson5910

2. jim_thompson5910

this sounds familiar...

Hmm lets see....

4. jim_thompson5910

we did this one?

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.

We did these :0

7. jim_thompson5910

oh gotcha

good memory :D.

I'm going with B!

10. itsssjennaaa

B

11. KG1234

b

12. jim_thompson5910

nope, not B

13. KG1234

kay dont hate me

14. itsssjennaaa

dang it! i thought i had it

15. itsssjennaaa

sowwwyy

Hi team, thanks for the backup!

17. itsssjennaaa

oh for sure!(;

18. KG1234

you got it kboyyy

See you later in chat QT pies.

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

or is this the confidence level BS...?

22. KG1234

im never going back on chat EVER

23. itsssjennaaa

me either!!!

NormalCfl(Wolfram).

25. itsssjennaaa

we got kicked off... hahahahha

26. KG1234

sooooo BA

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

28. KG1234

i told someone to die

29. 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)

So It's c....

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

Okay team onto the next Q!

33. itsssjennaaa

okay!