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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 185
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
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 185 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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jim_thompson5910Best ResponseYou've already chosen the best response.0
this sounds familiar...
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.0
we did this one?
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
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 185 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 185 The Pvalue 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

itsssjennaaaBest ResponseYou've already chosen the best response.0
dang it! i thought i had it
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
Hi team, thanks for the backup!
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
See you later in chat QT pies.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
Ok back to this stupid stats.... We gotta do some P<Z stuff right?
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
or is this the confidence level BS...?
 one year ago

KG1234Best ResponseYou've already chosen the best response.0
im never going back on chat EVER
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
NormalCfl(Wolfram).
 one year ago

itsssjennaaaBest ResponseYou've already chosen the best response.0
we got kicked off... hahahahha
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
You're cray... What did you say?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.0
sample proportion: m = x/n = 615/1205 = 0.510373 Margin of error: E = C*sqrt((m*(1m))/n) E = 2.0537489*sqrt((0.510373*(10.510373))/1205) E = 2.0537489*sqrt((0.510373*(10.510373))/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

KonradZuseBest ResponseYou've already chosen the best response.0
I'm going to write this formula down so I remember it...
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
Okay team onto the next Q!
 one year ago
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