iPad 7:07 PM * 49%-> ASSIGNMENT RESOURCES Chapter 4, Problem 137 Dr. Pittard Fal
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iPad 7:07 PM * 49%-> ASSIGNMENT RESOURCES Chapter 4, Problem 137 Dr. Pittard Fall2017 Hws Consider the Minitab output shown below One-Sample T: X Test of mu = 44.5 vs > 44.5 95% SE Lower Chapter4,Problem 123 VariableNMean StDev Mean Bound T P 16 45.8971 1.8273 45.0962 0.004 Revie Results by Study Objective (a) What is SE Mean? Round your answer to 3 decimal places. (b) What is T-value? Round your answer to 3 decimal places. (c) At what level of significance can the null hypothesis be rejected? Round your answer to 3 decimal places (d) If the hypotheses had been Ho: = 44 versus H1: > 44, would the P-value be larger (enter a value of 1) or smaller (enter a value of 2)? (e) If the hypotheses had been Hoi 44.5 versus H1: * 44.5, would you reject the null hypothesis at the 0.05 level (enter a value of 1 to reject, enter a value of 2 to accept)? LINK TO TEXTExplanation / Answer
Given that,
population mean(u)=44.5
sample mean, x =45.8971
standard deviation, s =1.8273
number (n)=16
null, Ho: =44.5
alternate, H1: >44.5
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.7531
since our test is right-tailed
reject Ho, if to > 1.7531
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 1.8273/ sqrt ( 16) )
= 0.457
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =45.8971-44.5/(1.8273/sqrt(16))
to =3.058
| to | =3.058
critical value
the value of |t | with n-1 = 15 d.f is 1.7531
we got |to| =3.058 & | t | =1.7531
make decision
hence value of | to | > | t | and here we reject Ho
p-value :right tail - Ha : ( p > 3.0583 ) = 0.00398
hence value of p0.05 > 0.00398,here we reject Ho
ANSWERS
---------------
a.
se = 0.457
b.
t = test statistic: 3.058
d.
p-value: 0.00398,
e. When two tailed at 0.05,
critical value
the value of |t | with n-1 = 15 d.f is 2.1314
we got |to| =3.058 & | t | =2.1314
make decision
hence value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 3.0583 ) = 0.008
value of p0.05 > 0.008,here we reject Ho
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