In: Statistics and Probability
Consider the following hypotheses:
H0: μ = 380
HA: μ ≠ 380
The population is normally distributed with a population standard
deviation of 63.
a-1. Calculate the value of the test statistic with x−x− = 393 and n = 95. (Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.)
a-2. What is the conclusion at the 10% significance level?
a-3. Interpret the results at αα = 0.10.
b-1. Calculate the value of the test statistic with x−x− = 359 and n = 95. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.)
b-2. What is the conclusion at the 5% significance level?
b-3. Interpret the results at αα = 0.05.
Part a-1)
Test Statistic :-
Z = ( X̅ - µ ) / ( σ / √(n))
Z = ( 393 - 380 ) / ( 63 / √( 95 ))
Z = 2.01
Part a-2)
Test Criteria :-
Reject null hypothesis if | Z | > Z( α/2 )
Critical value Z(α/2) = Z( 0.1 /2 ) = 1.645
| Z | > Z( α/2 ) = 2.0112 > 1.645
Result :- Reject null hypothesis
Part a-3)
There is sufficient evidence to support the claim that μ ≠ 380 at 10% level of significance.
part b-1)
Test Statistic :-
Z = ( X̅ - µ ) / ( σ / √(n))
Z = ( 359 - 380 ) / ( 63 / √( 95 ))
Z = -3.25
Part b-2)
Test Criteria :-
Reject null hypothesis if | Z | > Z( α/2 )
Critical value Z(α/2) = Z( 0.05 /2 ) = 1.96
| Z | > Z( α/2 ) = 3.2489 > 1.96
Result :- Reject null hypothesis
Part b-3)
There is sufficient evidence to support the claim that μ ≠ 380 at 5% level of significance.