Save -1.96 -2.17 1.64 -0.50 The degree of freedom for the t distribution is 19 1
ID: 3180310 • Letter: S
Question
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-1.96
-2.17
1.64
-0.50
The degree of freedom for the t distribution is
19
18
17
The p-value of the test statistics should be calculated using which function below?
=t.dist(t-stat, df, 1)
=t.dist.rt(t-stat, df)
=t.dist.2t(t-stat, df)
Let the p-value be 0.022. At a .05 level of significance, it can be concluded that the mean of the population is
significantly less than 6.975
not significantly less than 8
significantly greater than 8
significantly less than 8
a.-1.96
b.-2.17
c.1.64
d.-0.50
Explanation / Answer
Given that,
population mean(u)=8
sample mean, x =6.975
standard deviation, s =2
number (n)=18
null, Ho: >=8
alternate, H1: <8
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.74
since our test is left-tailed
reject Ho, if to < -1.74
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =6.975-8/(2/sqrt(18))
to =-2.174
| to | =2.174
critical value
the value of |t | with n-1 = 17 d.f is 1.74
we got |to| =2.174 & | t | =1.74
make decision
hence value of | to | > | t | and here we reject Ho
p-value :left tail - Ha : ( p < -2.1744 ) = 0.02204
hence value of p0.05 > 0.02204,here we reject Ho
ANSWERS
---------------
null, Ho: >=8
alternate, H1: <8
test statistic: -2.174
critical value: -1.74
decision: reject Ho
p-value: 0.02204
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