In: Statistics and Probability
Question No 4
Left -tailed test
z = -1.52
α = 0.04
Solution :-
Given :-
z = - 1.52 ( Left tailed test ) , = 0.04
By using Standard Normal ' z ' table
P ( z < -1.52 ) = 0.0643
P - Value = 0.0643
Since P- Value ( 0.0643 ) > ( 0.04 )
Decision : - Fail to reject Ho.
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Solution :-
Given :-
Mean (µ ) = 35000
Sample Mean ( x bar ) = 34450
Sample Standard deviation ( s ) = 1200
Sample Size ( n ) = 18
Significance level (α ) = 0.05
Df = n – 1 ; Df = 18 - 1 ; Df = 17
Hypothesis : -
Ho : µ = 35000
H1 : µ < 35000 ( This is Left tailed test )
Test statistic :-
t = ( x bar – µ ) / ( s / √ n )
t = ( 34450 - 35000 ) / ( 1200 / √ 18 )
t = - 1.94
t critical :- By using ‘ t ‘ distribution table at α = 0.05 and df = 17
t critical ( tc ) = - 1.740
P – value :- By using P value approach,P value at t = - 1.94 and df = 17
P – value = 0.0346
Decision :- Since P Value (0.0346 ) < α ( 0.05 )
Reject Ho.
Conclusion :- Therefore there is enough evidence to claim that population Mean is Less than 35000 at α = 0.05.