Question

Employees in a large company are entitled to 15-minute coffee breaks. A random sample of the duration of coffee breaks for 10 employees was taken with the times shown below. Assuming that the times are normally distributed, is there enough evidence at the 5% significance level to indicate that on average employees are taking longer coffee breaks than they are entitled to?

Data Collected
n	x
1	12
2	16
3	14
4	18
5	21
6	17
7	19
8	15
9	18
10	16

Q

1.The sample mean = ?

2. The population mean =?

3.Alpha Error = ?

4. The Confidience coefficient=

5.The Sample Size = ?

6.The sample Variance =?

7.The null hyposis is =

8. The alternative hyposis is =

9. the calculated test statistics( four decimale places) =

10.Rejection region (four decimal places) =

11. what do we conclude=

12. Write the conclusion so management can understand your finding?

Answer 1

Solution:

Given:

1.The sample mean =  Σx / n

=  12+16+14+18+21+17+19+15+18+16 / 10

= 166/ 10

sample mean = = 16.6

2. The population mean =  15

3.Alpha Error = 5% = 0.05

4. The Confidience coefficient = 1- Alpha Error = 0.95

5.The Sample Size = 10

6.The sample Variance = s2 =  Σ(xi - x̄)2 / n-1

= (12 - 16.6)2 + ... + (16 - 16.6)2  / 10-1

= 60.4 / 9

s2 = 6.71

s = 2.59

7.The null hyposis is = 15

8. The alternative hyposis is = 15

9. the calculated test statistics

t = ( - ) / (s /n)

t = (16.6 - 15) / ( 2.59 / 10 )

t = 1.9535

10.Rejection region = P-value < 0.05 then reject null hypothesis.

P-value = 0.4125

11. what do we conclude= Reject null hypothesis

12. There is sufficient evidence to support the on average employees are taking longer coffee breaks than they are entitled to  15-minute