Question

The mean number of sick days an employee takes per year is believed to be about 10. Members of a personnel department do not believe this figure. They randomly survey 8 employees. The number of sick days they took for the past year are as follows: 10; 6; 14; 4; 10; 9; 8; 9. Let X = the number of sick days they took for the past year. Should the personnel team believe that the mean number is about 10? Conduct a hypothesis test at the 5% level.

Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your answers to three decimal places.)

Answer 1

	Values ( X )	Σ ( Xi - X̅ )2
	10	1.5625
	6	7.5625
	14	27.5625
	4	22.5625
	10	1.5625
	9	0.0625
	8	0.5625
	9	0.0625
Total	70	61.5

Mean X̅ = Σ Xi / n
X̅ = 70 / 8 = 8.75
Sample Standard deviation SX = √ ( (Xi - X̅ )2 / n - 1 )
SX = √ ( 61.5 / 8 -1 ) = 2.9641

To Test :-

H0 :- µ = 10
H1 :- µ ≠ 10

Test Statistic :-
t = ( X̅ - µ ) / ( S / √(n))
t = ( 8.75 - 10 ) / ( 2.9641 / √(8) )
t = -1.1928

Test Criteria :-
Reject null hypothesis if | t | > t(α/2, n-1)
Critical value t(α/2, n-1) = t(0.05 /2, 8-1) = 2.365
| t | > t(α/2, n-1) = 1.1928 < 2.365
Result :- Fail to reject null hypothesis

There is insufficient evidence to support the claim that the mean number is about 10? Conduct a hypothesis test at the 5% level.

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 8- 1 ) = 2.365
8.75 ± t(0.05/2, 8 -1) * 2.9641/√(8)
Lower Limit = 8.75 - t(0.05/2, 8 -1) 2.9641/√(8)
Lower Limit = 6.272
Upper Limit = 8.75 + t(0.05/2, 8 -1) 2.9641/√(8)
Upper Limit = 11.228
95% Confidence interval is ( 6.272 , 11.228 )