show some work in support of your answers to receive full credit. Do not simply

Question

show some work in support of your answers to receive full credit. Do not simply report your final answer. 1. A civ il engineer is analyzing the compressive strength of concrete. Compressive strength is approximately Normally distributed with variance 2-1000 psi. 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 Ho: 3200 against Ha : > 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 neef 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, = sqrt(1000) = 31.6228

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/(31.6228/sqrt(12)

zo = 2.46475

| zo | = 2.46475

critical value

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

we got |zo| =2.46475 & | z | = 1.645

make decision

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

p-value : right tail - ha : ( p > 2.46475 ) = 0.00686

hence value of p0.05 > 0.00686, here we reject Ho

ANSWERS

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

null, Ho: =3200

alternate, H1: >3200

test statistic: 2.46475

critical value: 1.645

decision: reject Ho

p-value: 0.00686

Given that,

Standard deviation, =31.6228

Sample Mean, X =3222.5

Null, H0: =3200

Alternate, H1: !=3200

Level of significance, = 0.05

From Standard normal table, Z /2 =1.96

Since our test is two-tailed

Reject Ho, if Zo < -1.96 OR if Zo > 1.96

Reject Ho if (x-3200)/31.6228/(n) < -1.96 OR if (x-3200)/31.6228/(n) > 1.96

Reject Ho if x < 3200-61.980688/(n) OR if x > 3200-61.980688/(n)

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

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

becomes,

Reject Ho if x < 3200-61.980688/(12) OR if x > 3200+61.980688/(12)

Reject Ho if x < 3182.10771655 OR if x > 3217.89228345

Implies, don't reject Ho if 3182.10771655 x 3217.89228345

Suppose the true mean is 3230

Probability of Type II error,

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

= P(3182.10771655 x 3217.89228345 | 1 = 3230)

= P(3182.10771655-3230/31.6228/(12) x - / /n 3217.89228345-3230/31.6228/(12)

= P(-5.24633291 Z -1.32633291 )

= P( Z -1.32633291) - P( Z -5.24633291)

= 0.0924 - 0 [ Using Z Table ]

= 0.0924

For n =12 the probability of Type II error is 0.0924

POWER OF TEST = 1 - BETA = 1 - 0.0924 = 0.9076

PART C.

given that,

standard deviation, =31.6228

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 ( 31.6228/ Sqrt ( 12) ) ]

= [ 3222.5 - 1.96 * (9.129) , 3222.5 + 1.96 * (9.129) ]

= [ 3204.608,3240.392 ]

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) = 31.6228

ME =15

n = ( 1.96*31.6228/15) ^2

= (61.981/15 ) ^2

= 17.074 ~ 18

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.