An article in a journal reports that 34% of American fathers take no responsibility for child care. A researcher claims that the figure is higher for fathers in the town of Littleton. A random sample of 225 fathers from Littleton, yielded 97 who did not help with child care. Find the p-value for a test of the researcher's claim.

A.0.0015
B.0.0038
C.0.0529
D.0.0019

correct option is D.0.0019

Step-by-step explanation:

given data

journal reports p = 34%

random sample n = 225

yielded = 97

solution

we get here first Sample proportion that is

Sample proportion p1 =     ...............1

Sample proportion p1 = 0.4311

 and here Null hypothesis is

H0 =  p ≤ 0.34      ........................2

and

for Alternate hypothesis is

Ha =  p > 0.34     ..........................3

so we get here z statistics that is express as

z =         ....................4

put here value and we get

solve it we get

z = 2.88468

so here  p-value for a test  will be

P (z > 2.88468)

P = 1 - P (z < 2.88468)

P = 0.001957

so correct option is D.0.0019

1: -2 3/4, -2 1/2, -14, 0, 3, 8 1/3

2: -4, -3, -2 1/2, -2 1/4, 3 1/4,

step-by-step explanation: