Last year, television station WXYZ’s share of the 11 P.M. news audience was appr
ID: 3358931 • Letter: L
Question
Last year, television station WXYZ’s share of the 11 P.M. news audience was approximately equal to, but no greater than, 25 percent. The station’s management believes that the current audience share is higher than last year’s 25 percent share. In an attempt to substantiate this belief, the station surveyed a random sample of 400 11 P.M. news viewers and found that 146 watched WXYZ.
(a) Let p be the current proportion of all 11 P.M. news viewers who watch WXYZ. Set up the null and alternative hypotheses needed to attempt to provide evidence supporting the claim that the current audience share for WXYZ is higher than last year’s 25 percent share.
H0: p (Click to select)<> versus Ha: p (Click to select)><
Reject H0 at = (Click to select).10, .05, .01, .001.10 and .05.10.10, .05 and .01 ; (Click to select)Extremely strongVery strongNoSomeStrong evidence.
(c) Find the p-value for the hypothesis test in part b. Use the p-value to carry out the test by setting equal to .10, .05, .01, and .001. Interpret your results.
p-value is (Click to select)>< .001 ; (Click to select)Do not rejectReject H0 at all given values of
(d) Do you think that the result of the station’s survey has practical importance? (Round the sample proportion to 3 decimal places.)
The sample proportion of is significantly (Click to select)higherlower than .25.
Test of p = 0.25 vs p > 0.25 Sample X N Sample p Z-Value P-Value 1 146 400 0.365000 5.31 0.000Explanation / Answer
Given that,
possibile chances (x)=146
sample size(n)=400
success rate ( p )= x/n = 0.365
success probability,( po )=0.25
failure probability,( qo) = 0.75
null, Ho:p=0.25
alternate, H1: p>0.25
level of significance, = 0.01
from standard normal table,right tailed z /2 =2.33
since our test is right-tailed
reject Ho, if zo > 2.33
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.365-0.25/(sqrt(0.1875)/400)
zo =5.312
| zo | =5.312
critical value
the value of |z | at los 0.01% is 2.33
we got |zo| =5.312 & | z | =2.33
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 5.31162 ) = 0
hence value of p0.01 > 0,here we reject Ho
ANSWERS
---------------
a.
null, Ho:p=0.25
alternate, H1: p>0.25
test statistic: 5.312
b.
critical value: 2.33
decision: reject Ho
p-value: 0
c.
ii)at level of significance = 0.001
critical value
the value of |z | at los 0.001% is 3.09
we got |zo| =5.312 & | z | =3.09
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 5.31162 ) = 0
hence value of p0.001 > 0,here we reject Ho
iii) at level of significance = 0.05
| zo | =5.312
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =5.312 & | z | =1.64
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 5.31162 ) = 0
hence value of p0.05 > 0,here we reject Ho
iv) at level of significance = 0.10
zo =5.312
| zo | =5.312
critical value
the value of |z | at los 0.1% is 1.28
we got |zo| =5.312 & | z | =1.28
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 5.31162 ) = 0
hence value of p0.1 > 0,here we reject Ho
d.the result of the station’s survey has practical importance that
there is evidence that the current audience share is higher than last year’s 25 percent share
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