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 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 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.
 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 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.

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jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.0this sounds familiar...

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.0we did this one?

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0In 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.

itsssjennaaa
 2 years ago
Best ResponseYou've already chosen the best response.0dang it! i thought i had it

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0Hi team, thanks for the backup!

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0See you later in chat QT pies.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0Ok back to this stupid stats.... We gotta do some P<Z stuff right?

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0or is this the confidence level BS...?

KG1234
 2 years ago
Best ResponseYou've already chosen the best response.0im never going back on chat EVER

itsssjennaaa
 2 years ago
Best ResponseYou've already chosen the best response.0we got kicked off... hahahahha

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0You're cray... What did you say?

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.0sample 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)

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0I'm going to write this formula down so I remember it...

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0Okay team onto the next Q!
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