A civil engineer is analyzing the compressive strength of concrete. Compressive

Question

A civil engineer is analyzing the compressive strength of concrete. Compressive strength is approximately Normally distributed with variance ^2=1000 psi2. A random sample of n=12 specimens has a mean compressive strength of x =3222.5 psi. Let be the true mean compressive strength of all specimens from this type of concrete.

a)Test H_0:=3200 against H_a:>3200 at =0.05. Give the rejection region.

b)If the true value of is 3230 psi, what is the power of the test? Use =0.05.

c) Compute a 2-sided 95% CI for the mean compressive strength.

d)How large a sample does the engineer need so that a 2-sided 95% CI for will have a margin of error E of at least 15 psi?

Explanation / Answer

PART A.

Given that,

population mean(u)=3200

standard deviation, =31.62

sample mean, x =3222.5

number (n)=12

null, Ho: =3200

alternate, H1: >3200

level of significance, = 0.05

from standard normal table,right tailed z /2 =1.645

since our test is right-tailed

reject Ho, if zo > 1.645

we use test statistic (z) = x-u/(s.d/sqrt(n))

zo = 3222.5-3200/(31.62/sqrt(12)

zo = 2.46497

| zo | = 2.46497

critical value

the value of |z | at los 5% is 1.645

we got |zo| =2.46497 & | z | = 1.645

make decision

hence value of | zo | > | z | and here we reject Ho

p-value : right tail - ha : ( p > 2.46497 ) = 0.00685

hence value of p0.05 > 0.00685, here we reject Ho

ANSWERS

---------------

null, Ho: =3200

alternate, H1: >3200

test statistic: 2.46497

critical value: 1.645

decision: reject Ho

p-value: 0.00685

PART B.

Level of significance, = 0.05

From Standard normal table, Z /2 =1.645

Since our test is right-tailed

Reject Ho, if Zo < -1.645 OR if Zo > 1.645

Reject Ho if (x-3230)/31.62/(n) < -1.645 OR if (x-3230)/31.62/(n) > 1.645

Reject Ho if x < 3230-52.0149/(n) OR if x > 3230-52.0149/(n)

-----------------------------------------------------------------------------------------------------

Suppose the size of the sample is n = 12 then the critical region

becomes,

Reject Ho if x < 3230-52.0149/(12) OR if x > 3230+52.0149/(12)

Reject Ho if x < 3214.98459 OR if x > 3245.01541

Implies, don't reject Ho if 3214.98459 x 3245.01541

Suppose the true mean is 3230

Probability of Type II error,

P(Type II error) = P(Don't Reject Ho | H1 is true )

= P(3214.98459 x 3245.01541 | 1 = 3230)

= P(3214.98459-3230/31.62/(12) x - / /n 3245.01541-3230/31.62/(12)

= P(-1.645 Z 1.645 )

= P( Z 1.645) - P( Z -1.645)

= 0.95 - 0.05 [ Using Z Table ]

= 0.9

For n =12 the probability of Type II error is 0.9

Power of test = 1 - TYPE II ERROR = 1 -0.90 = 0.10

PART C.

TRADITIONAL METHOD

given that,

standard deviation, =31.62

sample mean, x =3222.5

population size (n)=12

I.

stanadard error = sd/ sqrt(n)

where,

sd = population standard deviation

n = population size

stanadard error = ( 31.62/ sqrt ( 12) )

= 9.128

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 * 9.128

= 17.891

III.

CI = x ± margin of error

confidence interval = [ 3222.5 ± 17.891 ]

= [ 3204.609,3240.391 ]

-----------------------------------------------------------------------------------------------

DIRECT METHOD

given that,

standard deviation, =31.62

sample mean, x =3222.5

population size (n)=12

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 = [ 3222.5 ± Z a/2 ( 31.62/ Sqrt ( 12) ) ]

= [ 3222.5 - 1.96 * (9.128) , 3222.5 + 1.96 * (9.128) ]

= [ 3204.609,3240.391 ]

-----------------------------------------------------------------------------------------------

interpretations:

1. we are 95% sure that the interval [3204.609 , 3240.391 ] 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 = 3222.5

standard error =9.128

z table value = 1.96

margin of error = 17.891

confidence interval = [ 3204.609 , 3240.391 ]

PART D.

Compute Sample Size  

n = (Z a/2 * S.D / ME ) ^2

Z/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )

Standard Deviation ( S.D) = 31.62

ME =15

n = ( 1.96*31.62/15) ^2

= (61.975/15 ) ^2

= 17.071 ~ 18

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