3. Your construction company is expanding an airport runway. You evaluate 28 soi

Question

3. Your construction company is expanding an airport runway. You evaluate 28 soil compaction tests. The average soil density is 122 Ibs't? with a standard deviation of 2.4 lbs/th. Using a t-test calculate a, 99% confidence interval b. 90%confidence interval 4. Your engineering firm is studying the effects of rainfall events in Norfolk neighborhoods prone to flooding. The existing flood maps indicate that a specific Norfolk neighborhood will experience flooding when the rainfall intensity is 0.65 inches hour. However, you are evaluating newer rainfall data and comparing it to the observed flooding. You specifically evaluate (35) rainfall events and conclude that flooding now occurs when the rainfall intensity is only 0.60 inches/hour. The standard deviation is 0.20 inches/hour. Using a z-test, develop Ho, Hi, and determine a p-value. Analyze and state your findings

Explanation / Answer

Q3.
a.
TRADITIONAL METHOD
given that,
sample mean, x =122
standard deviation, s =2.4
sample size, n =28
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 2.4/ sqrt ( 28) )
= 0.454
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 = 27 d.f is 2.771
margin of error = 2.771 * 0.454
= 1.257
III.
CI = x ± margin of error
confidence interval = [ 122 ± 1.257 ]
= [ 120.743 , 123.257 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =122
standard deviation, s =2.4
sample size, n =28
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 27 d.f is 2.771
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 = [ 122 ± t a/2 ( 2.4/ Sqrt ( 28) ]
= [ 122-(2.771 * 0.454) , 122+(2.771 * 0.454) ]
= [ 120.743 , 123.257 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 120.743 , 123.257 ] 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

b.
AT 90% CI
given that,
sample mean, x =122
standard deviation, s =2.4
sample size, n =28
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 27 d.f is 1.703
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 = [ 122 ± t a/2 ( 2.4/ Sqrt ( 28) ]
= [ 122-(1.703 * 0.454) , 122+(1.703 * 0.454) ]
= [ 121.228 , 122.772 ]

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