Question

According to MessageLabs Ltd., 80% of all email sent in July 2018 was spam. A system manager at a large corporation believes that the percentage at his company may be greater than 80%. He examines a random sample of 500 emails received at an email server and finds that 418 of the messages are spam.

a. Who/what is the population and what is the parameter of interest? Use the proper notation

b. What are the Null and Alternative hypotheses, in words and in notation? Ho: Ha:

c. This test is: One-sided Two-sided

d. Calculate the test statistic for our sample proportion, assuming the null hypothesis is true.

e. Are the inference conditions met in this situation for a one proportion z-test?

f. What is the probability that an email is spam in our sample, assuming the null hypothesis is true (i.e. compute the p-value)? (sketch helpful here)

g. Interpret the p-value

h. Based on the p-value, what is your conclusion regarding the Null Hypothesis? What is your conclusion in context?

i. What would the p-value be if we wanted to test if fewer than 80% of emails were spam at the manager’s company?

j. Would we expect a 95% confidence interval to include 0.80? Why or why not?

Answer 1

Given that,
possibile chances (x)=418
sample size(n)=500
success rate ( p )= x/n = 0.836
success probability,( po )=0.8
failure probability,( qo) = 0.2
null, Ho:p=0.8
alternate, H1: p>0.8
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=0.836-0.8/(sqrt(0.16)/500)
zo =2.012
| zo | =2.012
critical value
the value of |z α| at los 0.05% is 1.645
we got |zo| =2.012 & | z α | =1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: right tail - Ha : ( p > 2.01246 ) = 0.02209
hence value of p0.05 > 0.02209,here we reject Ho
ANSWERS
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a.
the population and the parameter of interest is Z test for proportion
b.
null, Ho:p=0.8
alternate, H1: p>0.8
c.
one sided test
d.
test statistic: 2.012
e.
yes,
the inference conditions met in this situation for a one proportion z-test
critical value: 1.645
decision: reject Ho
f.
the probability that an email is spam in our sample, assuming the null hypothesis is true
p-value: 0.02209
g.
p value is greater than alpha value
i.
the p-value be if we wanted to test if fewer than 80% of emails were spam at the manager’s compan
we have enough evidence to support the claim that A system manager at a large corporation believes that the percentage at his company may be greater than 80%.
j.
TRADITIONAL METHOD
given that,
possible chances (x)=418
sample size(n)=500
success rate ( p )= x/n = 0.836
I.
sample proportion = 0.836
standard error = Sqrt ( (0.836*0.164) /500) )
= 0.017
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.017
= 0.032
III.
CI = [ p ± margin of error ]
confidence interval = [0.836 ± 0.032]
= [ 0.804 , 0.868]
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DIRECT METHOD
given that,
possible chances (x)=418
sample size(n)=500
success rate ( p )= x/n = 0.836
CI = confidence interval
confidence interval = [ 0.836 ± 1.96 * Sqrt ( (0.836*0.164) /500) ) ]
= [0.836 - 1.96 * Sqrt ( (0.836*0.164) /500) , 0.836 + 1.96 * Sqrt ( (0.836*0.164) /500) ]
= [0.804 , 0.868]
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interpretations:
1. We are 95% sure that the interval [ 0.804 , 0.868] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion