Question

Define each of them and show the process of solving them with Excel.

What is critical value/standardized test statistic z/standardized test statistic t/P-value?

Answer 1

i.
critical value:
the critical value is set. so that the probability that the test statistic is beyond the critical value is at most equal to the significance level.
if the null hypothesis be true thanks. professor for defining a critical value.
ii.
test statistic:
A test statistic is a statistic (a quantity derived from the sample) used in statistical hypothesis testing.
A hypothesis test is typically specified in terms of a test statistic.
considered as a numerical summary of a data-set that reduces the data to one value
that can be used to perform the hypothesis test.
iii.
p value:
The p-value is defined as the probability, under the null hypothesis,
here simply denoted by (but is often denoted as opposed to which is sometimes used to represent the alternative hypothesis)
of obtaining a result equal to or more extreme than what was actually observed.

Assumed values given examples for Z test and t test for single means
process of test statistic and p value,critical values are shown in the examples.

a.
Given that,
population mean(u)=10.2
standard deviation, sigma =1.78
sample mean, x =10.9
number (n)=50
null, Ho: μ=10.2
alternate, H1: μ!=10.2
level of significance, alpha = 0.1
from standard normal table, two tailed z alpha/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) = x-u/(s.d/sqrt(n))
zo = 10.9-10.2/(1.78/sqrt(50)
zo = 2.78
| zo | = 2.78
critical value
the value of |z alpha| at los 10% is 1.645
we got |zo| =2.78 & | z alpha | = 1.645
make decision
hence value of | zo | > | z alpha| and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 2.78 ) = 0.01
hence value of p0.1 > 0.01, here we reject Ho
ANSWERS
---------------
null, Ho: μ=10.2
alternate, H1: μ!=10.2
test statistic: 2.78
critical value: -1.645 , 1.645
decision: reject Ho
p-value: 0.01
we have enough evidence to support the claim

b.
Given that,
population mean(u)=2.86
sample mean, x =2.8
standard deviation, s =0.85
number (n)=10
null, Ho: μ=2.86
alternate, H1: μ!=2.86
level of significance, alpha = 0.05
from standard normal table, two tailed t alpha/2 =2.262
since our test is two-tailed
reject Ho, if to < -2.262 OR if to > 2.262
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =2.8-2.86/(0.85/sqrt(10))
to =-0.223
| to | =0.223
critical value
the value of |t alpha| with n-1 = 9 d.f is 2.262
we got |to| =0.223 & | t alpha | =2.262
make decision
hence value of |to | < | t alpha | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -0.2232 ) = 0.8283
hence value of p0.05 < 0.8283,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=2.86
alternate, H1: μ!=2.86
test statistic: -0.223
critical value: -2.262 , 2.262
decision: do not reject Ho
p-value: 0.8283
we do not have enough evidence to support the claim