SAMPLE SIZE = 16
IF IN ADDITION TO THE SAMPLE SUMMARIES (SAMPLE MEAN) = $5,400 AND (SAMPLE STANDARD DEVIATION) = $1,280, THE POPULATION STANDARD DEVIATION IS KNOWN AS $1,250.
(A) AT THE 5% SIGNIFICANCE LEVEL, DO WE HAVE SUFFICIENT EVIDENCE THAT THE POPULATION AVERAGE BONUS WAS BELOW $6,000?
CIRCLE APPROPRIATE ANSWER: YES! NO!
(B) SHOW THE TEST STATISTIC VALUE, THE CRITICAL VALUE(S) NEEDED FOR YOUR DECISION AND FORMULATE THE REJECTION RULE.
TEST STATISTIC VALUE =
CRITICAL VALUE(S):
REJECTION RULE STATES...
(C) AT THE SAME 5% SIGNIFICANCE LEVEL, DO WE HAVE SUFFICIENT EVIDENCE THAT THE POPULATION AVERAGE BONUS EXCEEDED $4,900?
CIRCLE APPROPRIATE ANSWER: YES! NO!
(D) SHOW THE CRITICAL VALUE(S) NEEDED FOR YOUR DECISION AND FORMULATE THE REJECTION RULE.
TEST STATISTIC VALUE =
CRITICAL VALUE(S):
REJECTION RULE STATES...
1) H0: = 6000
H1: < 6000
The test statistic z = ()/(/)
= (5400 - 6000)/(1250/)
= -1.92
At = 0.05, the critical value is z0.05 = -1.645
Reject H0, if z < -1.645
Since the test statistic value is less than the critical value (-1.92 < -1.645), so we should reject the null hypothesis.
So at 5% significance level there is sufficient evidence that the population average bonus was below 6000
2) H0: = 4900
H1: > 4900
The test statistic z = ()/(/)
= (5400 - 4900)/(1250/)
= 1.6
At = 0.05, the critical value is z0.05 = 1.645
Reject H0, if z > 1.645
Since the test statistic value is not greater than the critical value (1.6 < 1.645), so we should not reject the null hypothesis.
So at 5% significance level there is not sufficient evidence that the population average bonus exceeded 4900.
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