(a) Calculate 95 percent and 99 percent confidence intervals for µ . (Round your
ID: 2921994 • Letter: #
Question
(a) Calculate 95 percent and 99 percent confidence intervals for µ. (Round your answers to 3 decimal places.)
(b) Using the 95 percent confidence interval, can we be 95 percent confident that µ is at least 50 pounds? Explain.
(Click to select)YesNo , 95 percent interval is (Click to select)abovebelow 50.
(c) Using the 99 percent confidence interval, can we be 99 percent confident that µ is at least 50 pounds? Explain.
(Click to select)YesNo , 99 percent interval extends (Click to select)belowabove 50.
(d) Based on your answers to parts b and c, how convinced are you that the new 30-gallon trash bag is the strongest such bag on the market?
(Click to select)NotFairly confident, since the 95 percent CI is (Click to select)abovebelow 50 while the 99 percent CI contains 50.
Explanation / Answer
(a) Calculate 95 percent and 99 percent confidence intervals for µ.
PART A.
TRADITIONAL METHOD
given that,
standard deviation, =1.68
sample mean, x =50.574
population size (n)=36
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 1.68/ sqrt ( 36) )
= 0.28
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 0.28
= 0.549
III.
CI = x ± margin of error
confidence interval = [ 50.574 ± 0.549 ]
= [ 50.025,51.123 ]
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DIRECT METHOD
given that,
standard deviation, =1.68
sample mean, x =50.574
population size (n)=36
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 50.574 ± Z a/2 ( 1.68/ Sqrt ( 36) ) ]
= [ 50.574 - 1.96 * (0.28) , 50.574 + 1.96 * (0.28) ]
= [ 50.025,51.123 ]
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interpretations:
1. we are 95% sure that the interval [50.025 , 51.123 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 50.574
standard error =0.28
z table value = 1.96
margin of error = 0.549
confidence interval = [ 50.025 , 51.123 ]
WHEN 99% CI
CI = confidence interval
confidence interval = [ 50.574 ± Z a/2 ( 1.68/ Sqrt ( 36) ) ]
= [ 50.574 - 2.576 * (0.28) , 50.574 + 2.576 * (0.28) ]
= [ 49.853,51.295 ]
(b) Using the 95 percent confidence interval, can we be 95 percent confident that µ is at least 50 pounds?
Yes, above, since interval achived is [ 50.025,51.123 ] is higher than 50
(c) Using the 99 percent confidence interval, can we be 99 percent confident that µ is at least 50 pounds?
no, below, since interval achived is [ 49.853,51.295 ] is lower to 50
(d) notFairly confident since the 95 percent CI is above 50 while the 99 percent CI contains 50.
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