Question

A SAMPLE OF 16 EMPLOYEES WAS DRAWN FOR THE ANALYSIS OF THE AVERAGE BONUS EARNED LAST YEAR. ASSUME THAT THE INDIVIDUAL BONUS IS NORMALLY DISTRIBUTED WITH UNKNOWN PARAMETERS. SAMPLE SUMMARIES ARE: (SAMPLE MEAN) = $5,400 AND (SAMPLE STANDARD DEVIATION) = $1,280.

(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...

Answer 1

1) H₀: = 6000

    H₁: < 6000

The test statistic t = ()/(s/)

                             = (5400 - 6000)/(1280/)

                             = -1.875

At = 0.05, the critical value is t_{0.05, 15} = -1.753

Reject H₀, if t < -1.753

Since the test statistic value is less than the critical value (-1.875 < -1.753), 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 t = ()/(s/)

                             = (5400 - 4900)/(1280/)

                             = 1.563

At = 0.05, the critical value is t_{0.05, 15} = 1.753

Reject H₀, if t > 1.753

Since the test statistic value is not greater than the critical value (1.563 < 1.753), 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.