I Really need help getting B &C, I believe A is 385<mean<401 Do you use a t tabl
ID: 3365531 • Letter: I
Question
I Really need help getting B &C, I believe A is 385<mean<401
Do you use a t table?? i dont know if it should be zscore or t
We destroy a number of concrete beams by applying a large tensile load. Each time, we write down the tensile strength of the beam from the test. We obtain the following data: 363 376 376 377 380 382 386 389 392 394 401 402 403 408 409 418 420 Assume that the tensile strength has a normal distribution, but we don't know the true mean or variance. We are curious about the average strength of the beams, and want to find a two-sided interval that we are 95% sure includes the true average. A)What is the interval? now assume that we're building a rather large bridge in an earthquake-prone region by using 500 of those beams. we add enough redundancy to the structure so that even if 10% of the beams fail in an earthquake, the bridge won't collapse. We want to find a design tensile strength such that, with 95% confidence, the bridge will remain standing after an earthquake. B) Use what you have learned about intervals to identify a design strength can we use, such that only 10% of the beams would be weaker than the chosen design strength. now imagine that we are building a rather small bridge using only one concrete beam, and we are curious about its strength. Obviously, we can't destroy it to find out, because then we can't use it for the bridge. C) What is the maximum tensile load we can apply to the beam if we want to be 99% confident that it will not fail?Explanation / Answer
a.
TRADITIONAL METHOD
given that,
sample mean, x =392.7059
standard deviation, s =16.143
sample size, n =17
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 16.143/ sqrt ( 17) )
= 3.915
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 16 d.f is 2.12
margin of error = 2.12 * 3.915
= 8.3
III.
CI = x ± margin of error
confidence interval = [ 392.7059 ± 8.3 ]
= [ 384.406 , 401.006 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =392.7059
standard deviation, s =16.143
sample size, n =17
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 16 d.f is 2.12
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 392.7059 ± t a/2 ( 16.143/ Sqrt ( 17) ]
= [ 392.7059-(2.12 * 3.915) , 392.7059+(2.12 * 3.915) ]
= [ 384.406 , 401.006 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 384.406 , 401.006 ] 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
b.
assume that we are biulding a rather larger bridge in an earth quake prone region by using 500 of those beams if 10% beams fail in an earth quake
the bridge won't collapse.95% of confidence the bridge will remain standing after each quake.the above interval 10% reduction design strengh
interval = (345.9654,360.9054)
c.
TRADITIONAL METHOD
given that,
sample mean, x =392.7059
standard deviation, s =16.143
sample size, n =17
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 16.143/ sqrt ( 17) )
= 3.915
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 16 d.f is 2.921
margin of error = 2.921 * 3.915
= 11.436
III.
CI = x ± margin of error
confidence interval = [ 392.7059 ± 11.436 ]
= [ 381.269 , 404.142 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =392.7059
standard deviation, s =16.143
sample size, n =17
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 16 d.f is 2.921
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 392.7059 ± t a/2 ( 16.143/ Sqrt ( 17) ]
= [ 392.7059-(2.921 * 3.915) , 392.7059+(2.921 * 3.915) ]
= [ 381.269 , 404.142 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 381.269 , 404.142 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.