A sample of 36 observations is selected from a normal population. The sample mea

Question

A sample of 36 observations is selected from a normal population. The sample mean is 21, and the population standard deviation is 5. Conduct the following test of hypothesis using the 0.05 significance level. H_0: mu lessthanorequalto 20 H_1: mu > 20 Is this a one- or two-tailed test? "One-tailed"- the alternate hypothesis is greater than direction. "Two-tailed" -the alternate hypothesis is different from direction. b. What is the decision rule? (Round your answer to 2 decimal places.) c. what is the value of the test statistic? (Round your answer to 2 decimal places.) d. What is your decision regarding H_0? Do not reject Reject There is evidence to conclude that the population mean is greater than 20. e. What is the p-value? (Round your answer to 4 decimal places.)

Explanation / Answer

Given that,
population mean(u)=20
standard deviation, =5
sample mean, x =21
number (n)=36
null, Ho: <=20
alternate, H1: >20
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 21-20/(5/sqrt(36)
zo = 1.2
| zo | = 1.2
critical value
the value of |z | at los 5% is 1.645
we got |zo| =1.2 & | z | = 1.645
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : right tail - ha : ( p > 1.2 ) = 0.11507
hence value of p0.05 < 0.11507, here we do not reject Ho

ANSWERS
---------------
a.
it is one tailed test

b.
since our test is right-tailed
reject Ho, if zo > 1.645

c.
test statistic: 1.2
null, Ho: <=20
alternate, H1: >20
test statistic: 1.2
critical value: 1.645
decision: do not reject Ho
p-value: 0.11507

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