Question

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

Answer 1

Answer:

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 = $\frac{97}{225}$     ...............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 =  $\frac{p1- p}{\sqrt{\frac{p(1-p)}{n}}}$       ....................4

put here value and we get

$z = \frac{0.4311-0.34}{\sqrt{\frac{0.34(1-0.34)}{225}}}$  

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