25) When blood levels are low at an area hospital, a call goes out to local resi

Question

25) When blood levels are low at an area hospital, a call goes out to local residents to give blood. The blood center is interested in determining which sex - males or females - is more likely to respond Random, independent samples of 60 females and 100 males were each asked if they would be willing to give blood when called by a local hospital. A success is defined as a person who responds to the call and donates blood. The goal is to compare the percentage of the successes of the male and female responses. Suppose 45 of the females and 60 of the males responded that they were able to give blood. Find the test statistic that would be used if it is desired to test to determine if a difference exists between the proportion of the females and males who responds to the call to donate blood A) z = 2.01 B) z = 1.645 C) z = 1.93 D)2 = 1.96

Explanation / Answer

This is case of difference in proportion test

here

p1 = 45/60 = 0.75
n1 = 60

p2 = 60/100 = 0.6
n2 = 100

so the z stats is
z = (p1 - p2) / SE

and SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

and p = (p1 * n1 + p2 * n2) / (n1 + n2)

so p =

(0.75 * 60 + 0.6 * 100) / (60 + 100) = 0.656

now standard error is

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
sqrt( 0.656 * ( 1 - 0.656 ) * [ (1/60) + (1/100) ] )
= 0.07757

and the z stat is thus

Z = (0.75-0.6)/0.07757 = 1.9337 ~ 1.93

Hence C

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