Question

1. A beverage company uses a machine to fill one-litre bottles with water. The machine is not perfect, so the amount of water delivered to each bottle is not exactly the same. The company wishes to estimte the mean volume of water delivered by the machine using a 90% confidence interval. To do so the company randomly samples 100 bottles, and measures the amount of water in each bottle. The mean volume of water in the sampled bottles was 998.9 mL. You can assume that the population standard deviation is 5.1 mL.

(a) (i) State the appropriate value of k for determining a 90% confidence interval.

k =......... .   (to 3 decimal places)

      (ii) Use the k value from part (i) and determine a 90% confidence interval for the true mean volume of water in the bottle using the formula from Lecture 5. Do not include the unit in the answer cells.

mL.......... < μ < ......... mL.      (to 3 decimal places)

Note: Since the sample size is considered sufficiently large (i.e. n > 30), it is acceptable to use the sample standard deviation s as an approximation to the population standard deviation σ in the confidence interval formula.

(b) What is the estimation error E of the 90% confidence interval calculated in part (a) (ii)?  Do not include the unit in the answer cells.

E = .......... mL.      (to 3 decimal places)

(c) Determine the sample size n necessary for the 90% confidence interval to have an estimation error below 0.5 mL. Remember to round up to the nearest whole number.

n = .      (exact value)

Answer 1

n= 100, c= 90%,  = 998.9, = 5.1

a)

i)

find k which  is the z critical value for c =  90%

using normal z table we get k as follows

K = 1.645

ii)

formula for confidence interval is

998.061 <    < 999.739

998.061 <    < 999.739

b)

formula to estimate error is

E = 0.83895

E = 0.839

c)

formula to calculate sample size is

n = 281.534841

n = 282