Statistical Inference

1. Dr. Abdul Aziz, Ph.D. in Business
Administration from University of Sindh.
abdul.aziz@fuuast.edu.pk

2. Dr. Abdul Aziz, Ph.D. in Business
Administration from University of Sindh.
abdul.aziz@fuuast.edu.pk

3. The Sampling Distribution of Sampling Mean
Handout Three-Chapter-7
(Practice Problems)
Dr. Abdul Aziz, Ph.D. in Business
Administration from University of Sindh.
abdul.aziz@fuuast.edu.pk

4. Q-1: The following table gives the monthly salaries (in $1000s) of the
six officers of a company.
A. Calculate Population Mean of Salary. µ
Answer1:
A.
Officer A B C D E F
Salary 8 12 16 20 24 28
There are 15 possible samples of size 4 from the population of six officers. They
are listed in the first column of the following table.
Sample Salaries ẋ-Mean Sample Salaries ẋ-Mean
A, B, C, D 8, 12, 16, 20 14 A, C, E, F
A, B, C, E 8, 12, 16, 24 15 A, D, E, F
A, B, C, F 8, 12, 16, 28 16 B, C, D, E
A, B, D, E 8, 12, 20, 24 16 B, C, D, F
A, B, D, F 8, 12, 20, 28 17 B, C, E, F
A, B, E, F 8, 12, 24, 28 18 B, D, E, F
A, C, D, E C, D, E, F
A, C, D, F
B. Complete the second and third columns of the table.
C. Complete the dot plot for the sampling distribution of the sample mean for samples of
size 4. Locate the population mean on the graph.
D. Obtain the probability that the mean salary of a random sample of four officers will be
within 1 (i.e., $1000) of the population mean.

5. Q-1: The following table gives the monthly salaries (in
$1000s) of the six officers of a company (Cont:)
.
Answer1:
C. Refer table in previous slide we have calculated means of sample size 4 and
placed dots in following dot plot chart. Calculate remaining means and place on the
following chart. We have also placed population mean 18 (green) on chart.
D. Data of the Q-1 is given in thousand, the population mean of the data is 18
means 18000, we have to find probability of the salary within ±1000 or between
17000 to 19000. First find z- scores of the data, we need standard deviation for z-
score calculation. The standard deviation of the data is 7.483315 or 7483.315. The z-
scores are -0.134 and +0.134 calculated below and areas of -0.134 is 0.4880 and
0.134 is 0.5517, subtract 0.4880 from 0.5517 we find 0.0637 or 6.37%. The
probability that salary will lie between 17000 to 19000 is 6.37%(Shaded)

6. Q-2: Suppose the speed limits in 13 countries in miles per hour are
as follows:
Answer2:
A. Frequency distribution is summarized in the
table and with the help of these calculations
mean is 68.01, and standard deviation is 9.4.
For mode and median calculations, use
previous concepts discussed in first lecture
“An Overview of Descriptive Statistics”.
Country Italy France Hungary Belgium Portugal Great Britain Spain Denmark Netherlands Greece Japan Norway Turkey
Highway Miles 87 81 75 75 75 70 62 62 62 62 62 56 56
A. What is the mean, standard deviation, median and mode for these
data? Prepare a frequency distribution.
B. Also calculate Z-scores for the data.
x f fx
56 2 112 291.70414
62 5 310 184.64497
70 1 70 3.69822
75 3 225 143.78698
81 1 81 167.00592
87 1 87 358.08284
13 885 1148.92308
x Z-score
56 -1.28461.650307981
56 -1.28461.650307981
62 -0.64640.417849491
62 -0.64640.417849491
62 -0.64640.417849491
62 -0.64640.417849491
62 -0.64640.417849491
70 0.20460.041845206
75 0.73640.542313873
75 0.73640.542313873
75 0.73640.542313873
81 1.3746 1.88966256
87 2.01294.051687199
885 0.0000 13
B. Use this formula to obtain z-score.
Mean of z-score is 0 and standard deviation of z-score is 1. It can be
confirmed with the help of table that sum of z-score is 0 when we
divide 0 with any number it would be 0. Similarly sum of is
13 and number of observations are also 13, when we divide it
would be 1 and square root of 1 is also 1. So standard deviation is
1. Remember mean of z score is always 0 and standard deviation is
always 1.

7. Q-3: In the year 2017, Pakistanis spent an average of Rs. 820,000
for a new car (light vehicle). Assume a standard deviation of Rs.
102,000.
A. Identify the population and variable under consideration.
B. For samples of 50 new car sales in 2017, determine the mean and
standard deviation of all possible sample mean prices.
C. Repeat part (B) for samples of size 100.
D. For samples of size 1000, answer the following question without
doing any computations: Will the standard deviation of all possible
sample mean prices be larger than, smaller than, or the same as that
in part (C)? Explain your answer
Answer3:
A. Population is Pakistani people and variable is spending on light vehicles.
B. There are three distributions:
1. Sample distribution: Where mean is denoted by and standard deviation is denoted by S.
2. Population distribution: Where mean is denoted by µ and standard deviation is denoted by σ.
3. Sampling distribution: A theoretical distribution which is used for hypothesis testing. The
mean of this distribution is same like or µ but standard deviation is denoted by .
We know mean of sampling distributions is same so mean of the data is Rs.
820,000 and standard deviation is which has been calculated
through formula given in sub part 3 of part B. Complete part C and D put
sample size n 100 and 1000 respectively and calculate standard deviation.

8. Dr. Abdul Aziz, Ph.D. in Business
Administration from University of Sindh.
abdul.aziz@fuuast.edu.pk
Q-4: According to the research, the mean number of hours (actually) worked by
female marketing and advertising managers is μ = 45 hours. Assuming a standard
deviation of σ = 7 hours, decide whether each of the following statements is true or
false or whether the information is insufficient to decide. Give a reason for each of
your answers.
A. For a random sample of 196 female marketing and advertising managers,
chances are roughly 95.44% that the sample mean number of hours worked will
be between 31 hours and 59 hours.
B. 95.44% of all possible observations of the number of hours worked by female
marketing and advertising managers lie between 31 hours and 59 hours.
C. For a random sample of 196 female marketing and advertising managers,
chances are roughly 95.44% that the sample mean number of hours worked will
be between 44 hours and 46 hours.
Answer4:
We know mean of sampling distribution is and
standard deviation is The mean in above example
is 45 and standard deviation is 7 when we convert it into standard deviation of
sampling distribution .
The z values at 95.44% confidence level are ±2, to calculate the range we use
= 44 to 46, option C is the
correct option.
,

9. Q-5: State Life Pakistan provides information about life insurance in force per
covered family in the Life Insurers Fact Book. Assume that the standard deviation of
life insurance in force is Rs. 50,900.
A. Determine the probability that the sampling error made in estimating the
population mean life insurance in force by that of a sample of 500 covered families
will be Rs. 2000 or less.
B. Must you assume that life-insurance amounts are normally distributed in order
to answer part (A)? What if the sample size is 20 instead of 500?
C. Repeat part (A) for a sample size of 5000.
Answer5:
The areas of z-score -0.88 and 0.88
are 0.1894 and 0.8106 respectively.
The probability of the sample mean is equal to population is 0.8106 –
0.1894 = 0.6212. There are 0.1894 or 18.94% chance it would be ≤ 2000.
Solve Part B and C of Q-5.

10. Q-6: A Berger company claims that its paint will last an average of 5
years. Assuming that paint life is normally distributed and has a
standard deviation of 0.5 year, answer the following questions:
A. Suppose that you paint one house with the paint and that the paint
lasts 4.5 years. Would you consider that evidence against the
manufacturer’s claim? (Hint: Assuming that the manufacturer’s claim
is correct, determine the probability that paint life for a randomly
selected house painted with the paint is 4.5 years or less.)
B. Suppose that you paint 10 houses with the paint and that the paint
lasts an average of 4.5 years for the 10 houses. Would you consider
that evidence against the manufacturer’s claim?
C. Repeat part (B) if the paint lasts an average of 4.9 years for the 10
houses painted.
Answer6:
Solve Q-6. Refer Q-5.

11. Q-7: A variable of a population is normally distributed with mean μ and standard
deviation σ. For samples of size n, fill in the blanks. Justify your answers.
A. 68.26% of all possible samples have means that lie within
________________________________________ of the population mean, μ.
B. 95.44% of all possible samples have means that lie within
________________________________________ of the population mean, μ.
C. 99.74% of all possible samples have means that lie within
________________________________________ of the population mean, μ.
Q-8: A variable of a population has mean μ and standard deviation σ.
For a large sample size n, fill in the blanks. Justify your answers.
A. Approximately __________________% of all possible samples
have means within σ/√n of the population mean, μ.
B. Approximately __________________% of all possible samples
have means within 2σ/√n of the population mean, μ.
C. Approximately __________________% of all possible samples have
means within 3σ/√n of the population mean, μ.
Fill the above blanks. Use values of Q-8 to fill blanks of Q-7, similarly use
values of Q-7 to fill the blanks of Q-8. Q-7 and Q-8 give same meaning.

12. Lynn Connaway
connawal@oclc.org
Thank YOU
Questions &
Suggestions
abdul.aziz@fuuast.edu.pk
abdulaziz2004@gmail.com
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