stat

stat assignment

7.9

ling should be taken to ensure that samples from each of the four strata will be proportionately included.

e.

select

0(

/

0)

5

50 invoices,

0/5000)=

100

0/5000)=

345

is not simple random sampling.

7.9 Solutio

n

a)

A stratified sam

p

b)

Here the number of sales invoices in each of the four strata differs significantly.

So the stratified sampling should be done by selecting samples proportional to the stratum si

z

For example,

from stratum Save Time On Research and Writing Hire a Pro to Write You a 100% Plagiarism-Free Paper. Get My Paper

1

5

0

50

500

=

invoices,

from stratum 2 select 500(500/5000) =

from stratum 3 select 500(

100

invoices, and

from stratum 4 select 500(

345

invoices to get a representative sample form each stratum.

c)

In simple random sampling, each unit in the population has an equal chance of being selected into the sample.

In stratified sampling, unlike the simple random sampling, proportionate representation across the entire population

is ensured by selecting samples from each stratum. That’s why the sampling in (

a)

7.21

and standard deviation 2/√25 = 2/5 = 0.4.

a)

and

minutes is given by,

z

7.8

=

8.2 0.5

=

.25

=

8 0

b)

=

=

=

c)

.

xbar z P(Z < z)

7.8 -1

8.2 1

=

=

=

7.21 Solution
Let Xbar denote the sample mean.
Then, Xbar is distributed as a normal distribution with mean	8
Therefore, Z = (Xbar – 8)/0.4 is distributed as a standard normal distribution.
The probability the sample mean is between		7.8		8.2		xbar				P(Z < z)
P(7.8 < Xbar < 8.2) = P[(7.8 - 8)/0.4 < Z < (8.2 - 8)/0.4]	–			0.5	0.3085
P(-0.5 < Z < 0.5)	0.6915
P(Z < 0.5) - P(Z < -0.5)	7.5		-1	0.1056
0.3829			0.5000
The probability that the sample mean is between 7.5 and 8 minutes is given by,
P(7.5 < Xbar < 8) = P[(7.5 - 8)/0.4 < Z < (8 - 8)/0.4]
P(-1.25 < Z < 0)
P(Z < 0) - P(Z < -1.25)
0.3944
If n = 100, Xbar is distributed as a normal distribution with mean 8 and standard deviation 2/√100 = 2/10 =		0.2
Therefore, Z = (Xbar – 8)/0.2 is distributed as a standard normal distribution.
Now, the probability the sample mean is between 7.8 and 8.2 minutes is given by,	0.1587
P(7.8 < Xbar < 8.2) = P[(7.8 - 8)/0.2 < Z < (8.2 - 8)/0.2]	0.8413
P(-1 < Z < 1)
P(Z < 1) - P(Z < -1)
0.6827
d)	When the sample size increases from n = 25 to n = 100, the standard error of the sample mean reduces from 0.4 to 0.2.
Therefore more samples will be closer to the distribution mean as the sample size increases from n = 25 to n = 100.
Hence the likelihood that the sample mean will fall within 0.2 minutes of the mean is much higher for samples of size 100 than for samples of size 25.

7.23

a)

b)

= 0.70

= √{0.70(1-0.70)/64}

=

7.23 Solution
It is given that, n = 64, x = 48
The sample proportion, p, of “successful” people is given by,
p = 48/64 =	0.75
It is given that, the population proportion,		π
	Now, the standard error of the proportion is given by,
		√{π(1-π)/n}
0.0573

7.25

a)

b)

Now, the standard error of the proportion is given by,

=

7.25 Solution
It is given that, n = 40,
x = Number of Y’s in the data = 14
The sample proportion, p, of college students who own shares of stock is given by,
p = 14/40 =	0.35
It is given that, the population proportion, π = 0.30
√{π(1-π)/n} = √{0.30(1-0.30)/40}
0.0725

7.29

n 100

π 0.25

√{π(1-π)/n} 0.0433

a)

=

p z P(Z < z)

= 0.5 0.25 0 0.5000
0.2

b)

=

= 0.1241 n 500
π 0.25
c)

√{π(1-π)/n} 0.0194

= P(Z ≤ 0) p z P(Z < z) = 0.5 0.25 0 0.5000

The probability that 20% or fewer will indicate that they have to pick up the slack for moms working flextime is given by, 0.2

=

= 0.0049

7.29 Solution
It is given that, n = 100, π =				0.25
Let p denote the sample proportion.
Thus, the sample proportion is normaly distributed with mean π =0.25 and standard deviation √{π(1-π)/n} =	0.0433
Therefore, Z = (p – 0.25)/0.0433 is distributed as Standard Normal.
The probability that 25% or fewer male employees will indicate that they have to pick up the slack for moms working flextime is given by,
P(p ≤ 0.25) = P[Z ≤ (0.25 – 0.25)/0.0433]
	P(Z ≤ 0)
-1.1547005384		0.1241
	The probability that 20% or fewer will indicate that they have to pick up the slack for moms working flextime is given by,
P(p ≤ 0.20) = P[Z ≤ (0.20 – 0.25)/0.0433]
P(Z ≤ -1.1547)
If n =500, the sample proportion is normaly distributed with mean π =0.25 and standard deviation √{π(1-π)/n} =	0.0194
Therefore, Z = (p – 0.25)/0.0194 is distributed as Standard Normal.
Now, the probability that 25% or fewer male employees will indicate that they have to pick up the slack for moms working flextime is given by,
P(p ≤ 0.25) = P[Z ≤ (0.25 – 0.25)/0.0194]
-2.5820		0.0049
P(p ≤ 0.20) = P[Z ≤ (0.20 – 0.25)/0.0194]
P(Z ≤ -2.5820)