i.com part2 code 395

                         

Course:

:

Level:

Statistics-II

I com 395 Assignment

   Assignment:   01

                       

Q. 1 (a) In a normal distribution 31 % of the items are under 54 and 8% are over 76.

Find mean and standard deviation of the distribution.

Solution

Ix = 76, n=54, - "x 76 x=~= 54=1.407

s2 = nIx² _(Ix)2

n( n-l)

_ 54(0.31)-(76)2

                        -       54(54-1)

16.74-5776

-

54(53)

_ -5759.26 2862

=~-2.01232 S=1.418

x+to ~=> 1.407+2.262x l~

                                  ~n                                    54

=> 1.407 +0.43649

=> 1.84349 > f1 >0.9751

(b) A r. v. X is normally distributed with mean 40 and standard deviation 4. (i) Find a point that has 97% of the distribution below it.

(ii) Find a point that has 62.2% of the distribution below it.

                       (iii)           Find a point that has 90% of the distribution above it.

(iv) Find two points containing the middle 98% area.

(v) Find two points containing middle 95% area.

              (i) Z = 0.97; then                      x=40+4x(0.97)=42.68

(ii) z = 0.622; then x=40+4x(0.622)=42.488

(iii) z = 0.9; then x=40+4x(O.9)=39.6(iv) z = 0.98' then x=40+4x(O.98)=43.12 (v) z = 0.95; then x=40+4x(O.95)=41.8

Q. 2 (a) Explain the following concepts:­(i) Population

(ii) Sample design (iii) Sampling error

(iv) Advantages of sampling (i) Population:

The word population in statistics is defined as the aggregate of object or totality of observation relating to under study problem. It means population does not mean living being only.

Statisticians define a population as the entire collection of items that is the focus of concern.

A population consists of an entire set of objects, observations, or scores that have something in common. For example, a population might be defined as all males between the ages of 15 and 18.

(ii) Sampling Design:

The description of the entire sampling unit Le. a computer list of the sampling unit is called sampling design.

In sample studies, we have to make a plan regarding the size of the sample, selection of the sample, collection of the sample data and preparation of the final results based on the sample study. The whole procedure involved is called the sample design. The term sample survey is used for a detailed study of the sample. In general, the term sample survey is used for any study conducted on the sample taken from some real world data.

(iii) Sampling Error

The standard deviation of the sample distribution of the mean ~, denoted by (J~; is called standard error.

These are the errors which occur due to the nature of sampling. The sample selected from the population is one of all possible samples. Any value calculated from the sample is based on the sample data and is called sample statistic. The sample statistic may or may not be close to the population parameter. If the statistic is

..•. ~

e and the true value of the population parameter is (), then the

.~

difference^e- .e is called sampling error. It is important to note that

a statistic is a random variable and it may take any value. A particular example of sampling error is the difference between the

sample mean^X and the population mean^Ji. Thus sampling error is also a random term. The population parameter is usually not known; therefore the sampling error is estimated from the sample data. The sampling error is due to the reason that a certain part of the population goes to the sample. Obviously, a part of the population cannot give the true picture of the properties of the population. But one should not get the impression that a sample always gives the result which is full of errors. We can design a sample and collect the sample data in a manner so that the sampling errors are reduced. The sampling errors can be reduced by the following methods: (1) by increasing the size of the sample (2) by stratification.

(iv) Advantages Of Sampling

Sampling has some advantages over the complete count. These are:

1. Need for Sampling:

Sometimes there is a need for sampling. Suppose we want to inspect the eggs, the bullets, the missiles and the tires of some firm. The study may be such that the objects are destroyed during the process of inspection. Obviously, we cannot afford to destroy all the eggs and the bullets etc. We have to take care that the wastage should be minimum. This is possible only in sample study. Thus sampling is essential when the units under study are destroyed.

2.      Saves Time and Cost:

As the size of the sample is small as compared to the population, the time and cost involved on sample study are much less than the Page30f3

complete counts. For complete count huge funds are required. There is always the problem of finances. A small sample can be studied in a limited time and total cost of sample study is very small. For complete count, we need a big team of supervisors and enumeration who are to be trained and they are to be paid properly for the work they do. Thus the sample study requires less time and less cost.

          3.      Reliability:

If we collect the information about all the units of population, the collected information may be. But we are never sure about it. We do not know whether the information is true or is completely false. Thus we cannot say anything with confidence about the quality of information. We say that the reliability is not possible. This is a very important advantage of sampling. The inference about the population parameters is possible only when the sample data is collected from the selected sample.

4. Sometimes the experiments are done on sample basis. The fertilizers, the seeds and the medicines are initially tested on samples and if found useful, then they are applied on large scale. Most of the research work is done on the samples.

          5.      Sample data is also used to check the accuracy of the census

data.

(b) A population consists of five units 3, 5, 7, 9 and 11. By taking a random sample of size 2 with replacement, verify the following properties

(i) Population mean is equal to the mean of the sample means.

(ii) Standard error of the sampling distribution of mean is less than the

population standard deviation.

Here population is 3,5,7,9 and 11. Population size = N = 5

Sample size = 2

No of possible sample = Nⁿ =5² =5x5=25

(All possible sample)

(3,3) (7,3) (11,3) (5,3) (9,3)

(3,7) (7,7) (11,7) (5,7) (9,7)

(3,11) (7,11) (11,11) (5,11) (9,11)

(3,5) (7,5) (11,5)

(5,5) (9,5)

(3,9) (7,9) (11,9) (5,9) (9,9)

(The mean of sample) 2 5 7 4 6 5 7 9 6 8

7        9        11 8            10

46857

6 8 10 7 9 Sampling distribution of ~

-	Tally	Frequency	Ix	ft2
x
2	1	1	2	4
5	111	3	15	5
7	11'1-1	5	35	245
9	111	3	27	243
11	1	1	11	121
8	1111	4	32	256
6	1111	4	24	144
10	11	2	20	200
4	11	2	8	32
		25	174	1320

Mean of Sampling distribution = fiX = 19= lZ:= 6.96 (ii) Standard Error:

x= Ifx²U If If

= 1320 _[_174J2

                                         54        54

= .)24.44-10.38 =3.749

Q. 3 (a) In a poll of 1000 voters selected at random from all the voters in a certain district, it is found that 518 voters are in favour of a particular candidate. Test the null hypothesis that the proportion of all the voters in the district who favour the candidate is equal to or less than 50% against the alternative that it is greater than 50% at a = 0.05.

Solution:

H :fi=1000 (i) Null hypothesis 0

HO:fi > 1000

(H) Level of significance =a = 0.05

X-fi Z= 0

          (Hi) Test of statistics           r-Jn

jx=40+4x(0

(iv) Calculation

518-1000

z= 0.51

lJ03

=518-1000=_68156

                   z       0.7072                .

(v) Critical region = is Z> 571.45 (vi) Conclusion

We reject H_O and conclude that the new method of voting has

increased the average vote.

(b) Distinguish between the following:

(i) Null and Alternative Hypothesis (ii) Acceptance and Rejection Region (iii) Type-I and Type-II Errors

(iv) Estimate and Estimator

(i) Null And Alternate Hvpothesis:

Null hypothesis is a hypothesis which is formulated and tested for possible rejection or nullification. A null hypothesis is usually denoted by symbolH_o' The hypothesis that the average height of Pakistani is 66" may be expressed below:

H_o : Ji = 66inches

Any hypothesis which differ from the null hypothesis is called as alternative hypothesis and is denoted by H_j or H_A• In a given test

there is usually only one null hypothesis but there may be many alternative hypotheses.

(H) Acceptance And Rejection Region:

Rejection region is the part of sampling distribution of a statistic which leads to rejection of the null hypothesis. The sample values beyond which H_o is rejected are called critical values. The part of

sampling distribution other then the critical region is called acceptance region.

All possible values which a test-statistic may assume can be divided into two mutually exclusive groups: one group consisting of values which appear to be consistent with the null hypothesis and the other having values which are unlikely to occur if Ho is true. The first group is called the acceptance region and the second set of values is known as the rejection region for a test. The rejection region is also called the critical region. The value(s) that separates the critical region from the acceptance region is called the critical value(s). The critical value which can be in the same units as the parameter or in the standardized units, is to be decided by the experimenter keeping in view the degree of confidence he (she) is willing to have in the null hypothesis.

(Hi) Type I And Type II Error:

A type I error, also known as a false positive, occurs when a statistical test rejects a true null hypothesis (HO). For example, if a null hypothesis states a patient is healthy, and the patient is indeed healthy, but the test rejects this hypothesis, falsely suggesting that the patient is sick. The rate of the type I error is denoted by the

Greek letter alpha (a) and usually equals the significance level of a test.

A type II error, also known as a false negative, occurs when the test fails to reject a false null hypothesis. For example, if a null hypothesis states a patient is healthy, and the patient is in fact sick, but the test fails to reject the hypothesis, falsely suggesting that the patient is healthy. The rate of the type II error is denoted by the Greek letter beta (P) and related to the power of a test (which equals I-P).

When we make a decision about the hypothesis our decision may be correct or incorrect. Generally the result of making such decisions can be distributed in the following way.

i. Two type of correct decision

ii. We reject a hypothesis which is actually true

iii. We accept a hypothesis which is actually not true.

(iv) Estimate Or Estimator:

The sample statistic which is used to estimate a population parameter is called an estimator and the numerical value of statistic is called estimate.

The estimate is the value obtained from a sample. The estimator, as used in statistics, is the method used. There's one more, the estimand, which is the population parameter. If we have an unbiased estimator, then after sampling many times, or with a large sample, we should have an estimate which is close to the estimand.

I will give you an example. I have a sample of 5 numbers and I take the average. The estimator is taking the average of the sample. It is the estimator of the mean of the population. The average = 4 (for example), this is my estimate.

An "estimator" or "point estimate" is a statistic (that is, a measurable function of the data) that is used to infer the value of an unknown parameter in a statistical model. The parameter being

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estimated is sometimes called the estimand. It can be either finite­dimensional (in parametric and semi-parametric models), or infinite-dimensional (semi-nonparametric and non-parametric models). If the parameter is denoted () then the estimator is typically written by adding a "hat" over the symbol: e. Being a function of the data, the estimator is itself a random variable; a particular realization of this random variable is called the "estimate". Sometimes the words "estimator" and "estimate" are used interchangeably.

Q. 4 (a) A sociologist gave a current events test to four blue-collar workers and four white-collar workers. The blue collar workers made scores of 23, 18, 22 and 21 and the white collar workers made scores of 17, 22, 19 and 18. Estimate the difference in mean scores for the two groups using a 99% confidence interval.

Solution(a)

         i.       Null hypothesis H 0 : f'I- f.l2 against

HI : f'I =t f.l₂

        H.       Level of significance = 0.99

       Hi.       Test statistic to be used

J-O

t= sd/

/.In

       iv.       We calculate the value of t from the data we denote the team program XI and X₂

		X2 -Xl = di	-	(d -dr
Xl	X2		d-d
23	17	-6	-4	16
18	22	4	6	36
22	19	-3	-1	1
21	18	-3	-1	1
---	---	-8	---	54

                               d-O        -2-0

t = sd/ = 4.24/ = 0.943

                       /.In       /-/4

v.        Critical Region:

t > to.025(3),

vi.       Conclusion:

Since the calculated value of 't' lies in the acceptance region, so we accept the null hypothesis.

   (b)       In a market survey a soft drink manufacturer wants to know whether there is a difference in product liking by two different age groups. In the younger group 70 out of 150 people liked the soft drink, while in the older group 38 out of 100 people like it. At a = 0.05 is there

For younger group:

70

~ =150,x₁ =70=>lJ =150=0.46=>q1 =l-lJ

= 1 - 0.46 = 0.54

For older group:

38

~ =100,x₁ =38=>lJ =100=0.38=>Q1 =l-lJ =1-0.38

= 0.62

Research Hypothesis:

Rate of soft drink is higher in older group rather than younger

lJ >P2 or lJ -P₂ >0

Testing Procedure:

  i.        HO:lJ -P₂ =>O=dOvsH A:lJ >P₂

ii.       Level of significance = 0.05

Hi.       Test statistic to be used:

Z= (P₁-P₂)-d_OP1Ql_ P₂^Q2

                       ~          n₂

Necessary Computation: 0.46-0.38-0

0.46x 0.54+ 0.38x 0.62

                                  150                100

                    Z=            0.08

-11.656+ 2.356

Z = 0.0399 Critical Region:

Zc >+za ~ Zc >+Z(0.05)

~ Zc >+ 1.645 but 0.50 < 1.645 So Zc lies in acceptance region

     vi.          Conclusion:

As Zc lies in acceptance region so we accept H_o• It means there data state do not provide different evidence.

Q. 5 (a) Two samples of seedlings were grown with 2 different fertilizers. The first sample, with 200 seedlings, had an average height of 10.9 inches and a standard deviation of 2.0 inches. The second sample with 100 seedlings, had an average height of 10.5 inches and a standard deviation of 5.0 inches. Do the data suggest that the difference between the average population heights is significant?

Solution

Groups	Average height	Standard	Size of sample
		deviation
Fertilizer	10.9	2.0	200
1	10.5	5.0	100

       i.          Null HypothesisH 0 : J1 - Ji₂

HI : Jil -:t- Ji2

     ii.          Level of significance a = 0.05

                                   10.9-10.5             0.4

z=-------~

0.402+ 25.25 2.821

                                   200        100

x² -x =di

1

(d_d)2

sl =2.0 s2 =5.0 n₁ =200 n₂ =100

                        s2 = n₂xs² = 100_x25=25.25                       __

                         1 n₂ -1 2 99                                         Z = Xl - x₂

                        s2 =25                                                       s2 s2

         2                       l+2

'1 '1

iv.          Computation:

                       

                         

'1 =200 n₂ =100

s2 = n₂x s2 = 100x 25 = 25.25 1 n₂ -1 2 99

                         10.9-10.5             0.4

z=-------~

0.402+ 25.25 2.821

                         200       100

Z = 0.14179

v.          Critical Region: Z < 0.123 and z > 1.96

vi.          Conclusion: As the value of Z lies in the critical region so we reject the null hypothesis. (b)       In a study of the effectiveness of physical exercise in weight reduction, a group of 16 persons engaged in a prescribed program of physical exercise for one month showed the following results (in pounds):

Perso	Weigh	Weigh	Perso	Weigh	Weigh
	t			t
n	Before	t After	n	Before	t After
1	209	196	9	170	164
2	178	171	10	153	152
3	169	170	11	183	179
4	212	207	12	165	162
5	180	177	13	201	199
6	192	190	14	179	173
7	158	159	15	243	231
8	180	180	16	144	140

Use the level of significance a = 0.01 to test the null hypothesis that the prescribed program of exercise is not effective (against a suitable alternative).

(b) Solution: i.       Null hypothesis HO :11- 112 HI :11* 112       ii.          Level of significance = 0.01      iii.          Test statistic to be used: d-O t= sd/ /~      iv.          C           t f

ompu a Ion:
Xl	x2	x2-x =di	d-d	(d _d)2
		I
209	196	-13	-9	81
178	171	-7	3	9
169	170	1	5	25
212	201	-5	-1	1

180	177	-3	1	1
192	190	-2	2	4
158	159	1	5	25
180	180	0	4	16
170	164	-6	-2	4
153	152	-1	3	9
183	179	-4	0	0
165	162	-3	1	1
201	199	-2	2	4
179	173	-6	-2	4
243	231	-12	-8	64
144	140	-4	0	0
---	----	-66	---	248

d = ~ I-di = =t!- = -4.125 Sd=~ nIlI(d_l-d)2 =~~~8 =15.5

t= d-O= -4.35-0=1 797

                                  sd/         7.041           .

/~ 1M

v.          Critical Region:

t > to.025(15)t > 2.262

vi.         Conclusion:

Since the calculated value of 't' lies in the acceptance region, so we accept the null hypothesis.