Question

In a random sample of 953 adults in a certain country, 18.4% of the sample members indicated some measure of disagreement with the statement "Globalization is more than an economic trade system instead it includes institutions and culture." a. Test at the 10% level the null hypothesis that at least 25% of all the country's adults would disagree with this statement. b. Find the probability of rejecting the null hypothesis with a 10%-level test if, in fact, 21% of all the country's adults would disagree with the statement. 5 Click the icon to view the standard normal distribution table. Determine the value for – ZQ, Za, or + 2a/2 as appropriate for this test. – 1.28 (Round to two decimal places as needed. Use a comma to separate answers as needed.) Calculate the test statistic. 6-Po -4.71 (Round to two decimal places as needed.) Po (1-P) /n State the conclusion of this test. Reject Ho. There is sufficient evidence to conclude that less than 25% of all the country's adults would disagree with this statement. b. The probability of rejecting the null hypothesis is (Round to four decimal places as needed.) when, in fact, 21% of all the country's adults would disagree with the statement.

Answer 1

a.
Given that,
possible chances (x)=175.352
sample size(n)=953
success rate ( p )= x/n = 0.18
success probability,( po )=0.25
failure probability,( qo) = 0.75
null, Ho:p=0.25
alternate, H1: p<0.25
level of significance, α = 0.1
from standard normal table,left tailed z α/2 =1.28
since our test is left-tailed
reject Ho, if zo < -1.28
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.184-0.25/(sqrt(0.1875)/953)
zo =-4.71
| zo | =4.71
critical value
the value of |z α| at los 0.1% is 1.28
we got |zo| =4.705 & | z α | =1.28
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: left tail - Ha : ( p < -4.70533 ) = 0
hence value of p0.1 > 0,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.25
alternate, H1: p<0.25
test statistic: -4.71
critical value: -1.28
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that less than 25% of all the country's adults
would disagree with this statement.
b.
Given that,
possible chances (x)=175.352
sample size(n)=953
success rate ( p )= x/n = 0.18
success probability,( po )=0.21
failure probability,( qo) = 0.79
null, Ho:p=0.21
alternate, H1: p!=0.21
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.64
since our test is two-tailed
reject Ho, if zo < -1.64 OR if zo > 1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.184-0.21/(sqrt(0.1659)/953)
zo =-1.97
| zo | =1.97
critical value
the value of |z α| at los 0.1% is 1.64
we got |zo| =1.971 & | z α | =1.64
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -1.97059 ) = 0.04877
hence value of p0.1 > 0.0488,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.21
alternate, H1: p!=0.21
test statistic: -1.97
critical value: -1.64 , 1.64
decision: reject Ho
p-value: 0.04877
we have enough evidence to support the claim that infact 21%of the country's adults would
disagree with the statement.