Exercise: The building specifications in a certain city require that the sewer p

Question

Exercise: The building specifications in a certain city require that the sewer pipe used in residential areas have a mean breaking strength of more than 2,500 pounds per lineal foot. A manufacturer who would like to supply the city with sewer pipe has submitted a bid and provided the following additional information Past experience indicates that the standard d strength is 115 pounds per lineal foot. 7 randomly selected sections of the manufacturer's pipe provided the average breaking strength of 2,571 pounds per lineal foot. (It is assumed that the breaking strength follows a normal distribution with parameters and .) a) Should the average breaking strength of the manufacturer's pipe be judged acceptable with eviation, , of the breaking type 1 error, = 0.01 ? b) c) What is the probability of failing to reject the null hypothesis at = 0.01 if the true breaking d) What sample size would be required to detect a true mean of 2,650 pounds per lineal foot if e) What is the smallest level of significance that would lead you to reject Ho in this test? strength of the manufacturer's pipe has a mean of 2,650 pounds per lineal foot? we wanted the power of the test to be at least 0.95? Find the 95% two-sided symmetrical confidence interval on the mean breaking strength.

Explanation / Answer

a.
Given that,
Standard deviation, =115
Sample Mean, X =2571
Null, H0: =2500
Alternate, H1: !=2500
Level of significance, = 0.01
From Standard normal table, Z /2 =2.58
Since our test is two-tailed
Reject Ho, if Zo < -2.58 OR if Zo > 2.58
Reject Ho if (x-2500)/115/(n) < -2.58 OR if (x-2500)/115/(n) > 2.58
Reject Ho if x < 2500-296.7/(n) OR if x > 2500-296.7/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 7 then the critical region
becomes,
Reject Ho if x < 2500-296.7/(7) OR if x > 2500+296.7/(7)
Reject Ho if x < 2387.858 OR if x > 2612.142
Suppose the true mean is 2571
Probability of Type I error,
P(Type I error) = P(Reject Ho | Ho is true )
= P(2387.858 < x OR x >2612.142 | 1 = 2571)
= P(2387.858-2571/115/(7) < x - / /n OR x - / /n >2612.142-2571/115/(7)
= P(-4.213 < Z OR Z >0.947 )
= P( Z <-4.213) + P( Z > 0.947)
= 0 + 0.1718 [ Using Z Table ]
= 0.172

b.
0.01 is level of significance that would lead you to reject Ho in the test

c.
Given that,
population mean(u)=2650
standard deviation, =115
sample mean, x =2571
number (n)=7
null, Ho: =2650
alternate, H1: !=2650
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
reject Ho, if zo < -2.576 OR if zo > 2.576
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 2571-2650/(115/sqrt(7)
zo = -1.81752
| zo | = 1.81752
critical value
the value of |z | at los 1% is 2.576
we got |zo| =1.81752 & | z | = 2.576
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -1.81752 ) = 0.06914
hence value of p0.01 < 0.06914, here we do not reject Ho
ANSWERS
---------------
null, Ho: =2650
alternate, H1: !=2650
test statistic: -1.81752
critical value: -2.576 , 2.576
decision: do not reject Ho
p-value: 0.06914
we failed to reject the null hypothesis at level of significance is 0.01

d.
Standard deviation, =115
Sample Mean, X =2571
Null, H0: =2500
Alternate, H1: !=2500
Level of significance, = 0.01
From Standard normal table, Z /2 =2.5758
Since our test is two-tailed
Reject Ho, if Zo < -2.5758 OR if Zo > 2.5758
Reject Ho if (x-2500)/115/(n) < -2.5758 OR if (x-2500)/115/(n) > 2.5758
Reject Ho if x < 2500-296.217/(n) OR if x > 2500-296.217/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = ? then the critical region
becomes,
Reject Ho if x < 2500-296.217/(7) OR if x > 2500+296.217/(7)
Reject Ho if x < 2388.0404977 OR if x > 2611.9595023
P(2388.0404977-2650/115/(n) x - / /n 2611.9595023-2650/115/(n)
power of the test = 1-beta = 0.95
beta = 1- 0.95 = 0.05 so that Z value = 1.96
2611.9595023-2650/115/(n)= 1.96
n =35.127 = 35

e.
TRADITIONAL METHOD
given that,
standard deviation, =115
sample mean, x =2571
population size (n)=7
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 115/ sqrt ( 7) )
= 43.466
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 * 43.466
= 85.193
III.
CI = x ± margin of error
confidence interval = [ 2571 ± 85.193 ]
= [ 2485.807,2656.193 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =115
sample mean, x =2571
population size (n)=7
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 = [ 2571 ± Z a/2 ( 115/ Sqrt ( 7) ) ]
= [ 2571 - 1.96 * (43.466) , 2571 + 1.96 * (43.466) ]
= [ 2485.807,2656.193 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [2485.807 , 2656.193 ] 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 = 2571
standard error =43.466
z table value = 1.96
margin of error = 85.193
confidence interval = [ 2485.807 , 2656.193 ]

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