HELPPP MEEE!!! Here is another question I got way more coming!!!!!!
2 answers:
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Answer:
Step-by-step explanation:
Given that:
(a) For x = 54 and s = 5.3
The test statistics can be computed as:
Z = -1.132
degree of freedom = n - 1
= 36 - 1
= 35
Using the Excel Formula:
P-Value = T.DIST(-1.132,35,1) = 0.1326
Decision: p-value is greater than significance level; do not reject
b
For x = 53 and s = 4.6
The test statistics can be computed as:
Z = -2.6087
degree of freedom = n - 1
= 36 - 1
= 35
Using the Excel Formula:
P-Value = T.DIST(-2.6087,35,1) =0.0066
Decision: p-value is < significance level; we reject the null hypothesis.
c)
For x = 56 and s = 5.0
The test statistics can be computed as:
Z = 1.2
degree of freedom = n - 1
= 36 - 1
= 35
Using the Excel Formula:
P-Value = T.DIST(1.2,35,1) = 0.88009
Decision: p-value is greater than significance level; do not reject