Question

In: Statistics and Probability

A researcher reported that 71.8% of all email sent in a recent month was spam. A...

A researcher reported that 71.8% of all email sent in a recent month was spam. A system manager at a large corporation believes that the percentage at his company may be 78%. He examines a random sample of 500 emails received at an email server, and finds that 367 of the messages are spam. Can you conclude that the percentage of emails that are spam differs from 78%? Use both α=0.05 and α=0.10 levels of significance and the P-value method with the TI-84 Plus calculator.

a. state null and alternative hypothesis. what tail is the hypothesis test?

b. Compute the value of the test statistic.

c. Using α=0.05, can you conclude that greater than 71% of emails are spam?

d. Using α=0.10, can you conclude that greater than 71% of emails are spam?

e. state conclusion

Solutions

Expert Solution

a.
Given that,
possibile chances (x)=367
sample size(n)=500
success rate ( p )= x/n = 0.734
success probability,( po )=0.78
failure probability,( qo) = 0.22
null, Ho:p=0.78
alternate, H1: p!=0.78
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
reject Ho, if zo < -1.645 OR if zo > 1.645
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.734-0.78/(sqrt(0.1716)/500)
zo =-2.483
| zo | =2.483
critical value
the value of |z α| at los 0.1% is 1.645
we got |zo| =2.483 & | z α | =1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -2.48304 ) = 0.01303
hence value of p0.1 > 0.013,here we reject Ho
ANSWERS
---------------
i.
null, Ho:p=0.78
alternate, H1: p!=0.78
test statistic: -2.483
critical value: -1.645 , 1.645
decision: reject Ho
p-value: 0.01303
we have enough evidence to support the claim that the percentage of emails that are spam differs from 78%
ii.
level of significance =0.05
Given that,
possibile chances (x)=367
sample size(n)=500
success rate ( p )= x/n = 0.734
success probability,( po )=0.78
failure probability,( qo) = 0.22
null, Ho:p=0.78
alternate, H1: p!=0.78
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=0.734-0.78/(sqrt(0.1716)/500)
zo =-2.483
| zo | =2.483
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =2.483 & | 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 != -2.48304 ) = 0.01303
hence value of p0.05 > 0.013,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.78
alternate, H1: p!=0.78
b.
test statistic: -2.483
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.01303
we have enough evidence to support the claim that the percentage of emails that are spam differs from 78%
c.
Given that,
possibile chances (x)=367
sample size(n)=500
success rate ( p )= x/n = 0.734
success probability,( po )=0.71
failure probability,( qo) = 0.29
null, Ho:p=0.71
alternate, H1: p>0.71
level of significance, α = 0.05
from standard normal table,right tailed z α/2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.734-0.71/(sqrt(0.2059)/500)
zo =1.1827
| zo | =1.1827
critical value
the value of |z α| at los 0.05% is 1.645
we got |zo| =1.183 & | z α | =1.645
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: right tail - Ha : ( p > 1.18268 ) = 0.11847
hence value of p0.05 < 0.11847,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.71
alternate, H1: p>0.71
test statistic: 1.1827
critical value: 1.645
decision: do not reject Ho
p-value: 0.11847
we do not have enough evidence to support the claim that greater than 71% of emails are spam
d.
Given that,
possibile chances (x)=367
sample size(n)=500
success rate ( p )= x/n = 0.734
success probability,( po )=0.71
failure probability,( qo) = 0.29
null, Ho:p=0.71
alternate, H1: p>0.71
level of significance, α = 0.1
from standard normal table,right tailed z α/2 =1.282
since our test is right-tailed
reject Ho, if zo > 1.282
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.734-0.71/(sqrt(0.2059)/500)
zo =1.1827
| zo | =1.1827
critical value
the value of |z α| at los 0.1% is 1.282
we got |zo| =1.183 & | z α | =1.282
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: right tail - Ha : ( p > 1.18268 ) = 0.11847
hence value of p0.1 < 0.11847,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.71
alternate, H1: p>0.71
test statistic: 1.1827
critical value: 1.282
decision: do not reject Ho
p-value: 0.11847
e.
we do not have enough evidence to support the claim that greater than 71% of emails are spam

orchestra answered 2 years ago
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