(06.04 LC) A children\'s cancer research fund has been set up to collect donatio

Question

(06.04 LC) A children's cancer research fund has been set up to collect donations for the treatment and research of the disease. A random sample of 600 people shows that 47% of the 300 who were contacted by telephone actually made contributions, compared with only 34% of the 300 who received email requests, which of the formulas calculates the 98% confidence interval or the difference in the proportions of people who make donations when contacted tele one ver sus se contacted by email? (4 points) 0.47)(0.34) (0.47)( 0.34) ( 0.47-0.34 ) ± 2.326 0.47)(0.53) (0.34)(0.66) , 300 ( 0.47-0.34 2.326 300 0.47)0.53) (0.34)(0.66) (0.47-0.34) 2.326 | 600 600 ( 0.47-0.34 )± 2.326,|(047)(0201 300 ,0a7 R 0.34) ( 0.47-0.34 2.326 600

Explanation / Answer

1)option 2nd is correct ,( 0.47-0.34) -/+ 2.326(0.47*0.53)/300+(0.34*0.66)/300

2)2nd option : includes zero ; conclude...may be the same

3)

4)

hence 99% CI =0.17 -/+ 0.096

5)

std error     Se =     (p1*(1-p1)/n1+p2*(1-p2)/n2)                                   = 0.0494

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