A soft-drink machine is regulated so that the amount of drink dispensed is approximately normally distributed with known standard deviation, sigma. Given a random sample of n drinks and the sample mean, x-bar, you find a 90% confidence interval for the mean of all drinks dispensed by this machine. Then you calculate a 90% confidence interval (same confidence level) using a larger sample, for example (n 20) drinks. Also, you notice that the sample mean, x-bar, is the same for both samples. Required:
Was this a reasonable decision?

This is  a reasonable decision because the sample size has no effect on the 90% confidence interval

Step-by-step explanation:

90% confidence interval

larger sample size = 20

condition : sample mean ( x-bar ) is the same for both samples

This is  a reasonable decision because the sample size has no effect on the 90% confidence interval

from condition 1 :

Amount of drink dispensed is normally distributed with known standard deviation , given a random sample of n drinks and the sample mean at a confidence interval of 90%

for condition 2 :

sample size = 20

mean = 2.25 ( assumed value )

std = 0.15 ( assumed value )

Z = 1.645 ( Z-value )

determine the 90% confidence interval

= mean ± z

= 2.25 ± 1.645

= 2.25 ± 0.0335 = ( 2.2835 , 2.2165 )

1. 3.4/4

2. 0.8/7

3. 0.5 x 3

4. 0.9 x 6

5. 10.8

6. 2.35

7. 0.567

8. 9.35

9. 0.02

10. 0.06

11. 0.16

12. 0.42

13. 0.63

14. 5.6

15. 8.1

16. 9.9

17. 0.28

18. 0.48

19. 0.27

20. 0.49

21. 0.45

22. 0.72

23. 0.9

24. 0.88

25. 0.75

26. 0.16

27. 0.328

28. 0.495

29. 0.72

30. 1.068

you have to put the graph on here with the question

step-by-step explanation: