Question

The capacities (in ampere-hours) were measured for a sample of 120 batteries. The average was 178 and the standard deviation was 15.

a)Find a 95% confidence interval for the mean capacity of batteries produced by this method. Round the answers to three decimal places.The 95% confidence interval is?

b) Find a 99% confidence interval for the mean capacity of batteries produced by this method. Round the answers to three decimal places.The 99% confidence interval is?

c) An engineer claims that the mean capacity is between 175.864 and 180.136 ampere-hours. With what level of confidence can this statement be made? Express the answer as a percent and round to two decimal places.

d) Approximately how many batteries must be sampled so that a 95% confidence interval will specify the mean to within ±2 ampere-hours? Round up the answer to the nearest integer.

e) Approximately how many batteries must be sampled so that a 99% confidence interval will specify the mean to within ±2 ampere-hours? Round up the answer to the nearest integer

please dont answer the quastion if you dont know the right answer.

Answer 1

a)

95% confidence interval is 175.316 ; 180.684

b)

for 99 % CI value of z=				2.580
margin of error E=z*std error                            =				3.533
lower confidence bound=sample mean-margin of error=				174.467
Upper confidence bound=sample mean +margin of error=				181.533

99% confidence interval is 174.467 ; 181.533

c)

here margin of error =(upper limit -lower limit)/2=(180.136-175.864)/2=2.136

therefore critical value z =margin of error/std error=2.136/1.369=1.56

from table 1.56 is critical value for 88% confidence interval

therefore level of confidence =88.00%

d)

for 95 % CI value of z=	1.960
standard deviation $\sigma$ =	15.00
margin of error E =	2
required sample size n=(z $\sigma$ /E)2    =	217.0

e)

for 99 % CI value of z=

2.580

required sample size n=(z $\sigma$ /E)2    =

375