Suppose a national survey indicated that 56% of women have had experience with a global positioning...

Suppose a national survey indicated that 56% of women have had experience with a global positioning system (GPS) device.

The survey indicated that 66% of the men surveyed have used a GPS device.

a. If the survey was based on a sample size of 200 men, do these data indicate that the proportion of men is the same as the proportion of women who have had experience with a GPS device? Use a significance level of 0.05

b. Obtain the p-value for the test indicated in part a.

Expert Solution

Given that,
possibile chances (x)=132
sample size(n)=200
a.
success rate ( p )= x/n = 0.66
success probability,( po )=0.56
failure probability,( qo) = 0.44
null, Ho:p=0.56
alternate, H1: p!=0.56
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.66-0.56/(sqrt(0.2464)/200)
zo =2.849
| zo | =2.849
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =2.849 & | z α | =1.96
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.84901 ) = 0.00439
hence value of p0.05 > 0.0044,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.56
alternate, H1: p!=0.56
test statistic: 2.849
critical value: -1.96 , 1.96
decision: reject Ho
b.
p-value: 0.00439
we have enough evidence to support the claim that national survey indicated that 56% of women have had experience with a global positioning.

orchestra answered 2 years ago

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