Unit  5
Correlation and regression analysis
Q1) Define correlation.
A1)
If the change in one variable affects a change in other variable, then these two variables are said to be correlated.
Positive correlation When both variables move in the same direction, or if the increase in one variable results in a corresponding increase in the other one is called positive correlation.
Negative correlation When one variable increases and other decreases or viceversa, then the variables said to be negatively correlated.
No correlation When two variables are independent and do not affect each other then there will be no correlation between the two and said to be uncorrelated.
Note (Perfect correlation) When a variable changes constantly with the other variable, then these two variables are said to be perfectly correlated.
Q2) What is Karl Pearson’s coefficient of correlation?
A2)
Karl Person’s coefficient of correlation is also called product moment correlation coefficient.
It is denoted by ‘r’, and defined as
Here are the standard deviations of these series.
Alternate formula
Note
1. Correlation coefficient always lies between 1 and +1.
2. Correlation coefficient is independent of change of origin and scale.
3. If the two variables are independent then correlation coefficient between them is zero.
Q3) Find the correlation coefficient between Age and weight of the following data
Age  30  44  45  43  34  44 
Weight  56  55  60  64  62  63 
A3)
x  y  ( ))  
30  56  10  100  4  16  40 
44  55  4  16  5  25  20 
45  60  5  25  0  0  0 
43  64  3  9  4  16  12 
34  62  6  36  2  4  12 
44  63  4  16  3  9  12 
Sum= 240 
360 
0 
202 
0 
70

32 
Karl Pearson’s coefficient of correlation
Q4) Find the correlation coefficient between the values X and Y of the dataset given below by using shortcut method
X  10  20  30  40  50 
Y  90  85  80  60  45 
A4)
X  Y  
10  90  20  400  20  400  400 
20  85  10  100  15  225  150 
30  80  0  0  10  100  0 
40  60  10  100  10  100  100 
50  45  20  400  25  625  500 
Sum = 150 
360 
0 
1000 
10 
1450 
1150 
Shortcut method to calculate correlation coefficient
Q5) Two variables X and Y are given in the dataset below, find the two lines of regression.
x  65  66  67  67  68  69  70  71 
y  66  68  65  69  74  73  72  70 
A5)
The two lines of regression can be expressed as
And
x  y  Xy  
65  66  4225  4356  4290 
66  68  4356  4624  4488 
67  65  4489  4225  4355 
67  69  4489  4761  4623 
68  74  4624  5476  5032 
69  73  4761  5329  5037 
70  72  4900  5184  5040 
71  70  5041  4900  4970 
Sum = 543  557  36885  38855  37835 
Now
And
Standard deviation of x
Similarly
Correlation coefficient
Put these values in regression line equation, we get
Regression line y on x
Regression line x on y
Q6) Derive Spearman rank correlation.
A6)
Spearman’s rank correlation
Let be the ranks of individuals corresponding to two characteristics.
Assuming nor two individuals are equal in either classification, each individual takes the values 1, 2, 3, and hence their arithmetic means are, each
Let , , , be the values of variable and , , those of
Then
Where and y are deviations from the mean.
Clearly, and
Q7) Compute the Spearman’s rank correlation coefficient of the dataset given below
Person  A  B  C  D  E  F  G  H  I  J 
Rank in test1  9  10  6  5  7  2  4  8  1  3 
Rank in test2  1  2  3  4  5  6  7  8  9  10 
A7)
Person  Rank in test1  Rank in test2  d =  
A  9  1  8  64 
B  10  2  8  64 
C  6  3  3  9 
D  5  4  1  1 
E  7  5  2  4 
F  2  6  4  16 
G  4  7  3  9 
H  8  8  0  0 
I  1  9  8  64 
J  3  10  7  49 
Sum 


 280 
Q8) What is likelihood function?
A8)
Let be a random sample of size n from a population with density function f (x. ). Then the likelihood function of the sample values . Usually denoted by L = L () is their joint density function; givenby
L gives the relative likelihood that the random variables assume a particular set of values For a given sample L becomes a function of the variable , the parameter.
Q9) Prove that the. Maximum likelihood estimates of the parameter a of a population having density function:
For a sample of, unit size is 2x. x being the sample value. Show also that the estimate is biased.
A9)
For a random sample of unit size (n = 1). The likelihood function is
Likelihood equation gives
Hence MLE of is given by
Since is not an unbiased estimate of
Q10) Write a short note on method of moments.
A10)
The principle of this method consists of equating the sample moments to the corresponding moments of the population, which are the function of unknown population parameter.
Here we will equate as many sample moments as there are unknown parameters and solve these simultaneous equations for estimating unknown parameter(s). This method of obtaining the estimate(s) of unknown parameter(s) is called “Method of Moments”.
Suppose be a random sample of size n taken from a population whose probability density (mass) function is f(x,) with k unknown parameters, say,
Then the r’th sample moment about origin is
And about mean is
While the rth population moment about origin is
And about mean is
Generally, the first moment about origin (zero) and rest central moments (about mean) are equated to the corresponding sample moments. Thus, the equations are
By solving these k equations for unknown parameters, we get the moment estimators.
Q11) Explain confidence intervals.
A11)
Letbe a random sample of size n taken from normal population N () when is known.
To find out the confidence interval for population mean, first of all we search the statistic for estimating μ whose distribution is known.
We know that when parent population is normal N () then sampling distribution of sample mean is normally distributed with mean and variance /n.
The variate
Follows the normal distribution with mean 0 and variance unity. Therefore, the probability density function of Z is
Here we will introduce two constants,
Where, is the value of the variate Z having an area of under the right tail of the probability curve of Z
By putting the value of Z in above equation, we get
Multiplying each term by (1) in above inequality, we get
This can be rewritten as
Hence, (1) 100% confidence interval of population mean is given by
Q12) The mean life of the bulbs manufactured by a company follows normal distribution with standard deviation 3200 hrs. A sample of 250 bulbs is taken and it is found that the average life of the bulbs is 50000 hrs with a standard deviation of 3500 hrs. Establish the 99% confidence interval within which the mean life of bulbs of the company is expected to lie.
A12)
Here, we have
n = 250 , and S = 3500
Since population standard deviation i.e., population variance is known,
Therefore, we use (1) 100% confidence limits for population mean when population variance is known which are given by
Where is the value of the variate Z having an area of under the right tail of the probability curve of Z and for 99% confidence interval, we have
Therefore, the 99% confidence limits are
By putting the values of n, and σ, the 99% confidence limits are
Hence, 99% confidence interval within which the mean life of bulb of the company is expected to lie is [49477.80, 50522.20].
Q13) What is pvalue?
A13)
The pvalue is the smallest value of level of significance (α) at which a null hypothesis can be rejected using the obtained value of the test statistic and it is defined as below
The pvalue is the probability of obtaining a test statistic equal to or more extreme (in the direction of sporting H1) than the actual value obtained when null hypothesis is true.
While testing a hypothesis, preselection of a significance level α does not account for values of test statistics that are “close” to the critical region.
Thus a test statistic value that is nonsignificant say for α = 0.05 may become significant for α = 0.01.
pvalue approach is designed to give the user an alternative (in terms of probability) to a mere “reject” or “do not reject” conclusion.
Q14) A set of five similar coins is tossed 320 times and the result is
Number of heads  0  1  2  3  4  5 
Frequency  6  27  72  112  71  32 
A14)
For v = 5, we have
P, probability of getting a head=1/2; q, probability of getting a tail=1/2.
Hence the theoretical frequencies of getting 0,1,2,3,4,5 heads are the successive terms of the binomial expansion
Thus, the theoretical frequencies are 10, 50, 100, 100, 50, 10.
Hence,
Since the calculated value of is much greater than the hypothesis that the data follow the binomial law is rejected.
Q15) In experiments of pea breeding, the following frequencies of seeds were obtained
Round and yellow  Wrinkled and yellow  Round and green  Wrinkled and green  Total 
316  101  108  32  556 
Theory predicts that the frequencies should be in proportions 9:3:3:1. Examine the correspondence between theory and experiment.
A15)
The corresponding frequencies are
Hence,
For v = 3, we have
Since the calculated value of is much less than there is a very high degree of agreement between theory and experiment.
Q16) Explain the testing of hypothesis.
A16)
Hypothesis
A hypothesis is a statement or a claim or an assumption about the value of a population parameter.
Similarly, in case of two or more populations a hypothesis is comparative statement or a claim or an assumption about the values of population parameters.
For example
If a customer of a car wants to test whether the claim of car of a certain brand gives the average of 30km/hr is true or false.
Simple and composite hypotheses
If a hypothesis specifies only one value or exact value of the population parameter then it is known as simple hypothesis. And if a hypothesis specifies not just one value but a range of values that the population parameter may assume is called a composite hypothesis.
Null and alternative hypothesis
The hypothesis which is to be tested as called the null hypothesis.
The hypothesis which complements to the null hypothesis is called alternative hypothesis.
In the example of car, the claim is and its complement is .
The null and alternative hypothesis can be formulated as
And
Q17) What are type1 and type2 errors?
A17)
Type1 error
The decision relating to rejection of null hypo. When it is true is called type1 error.
The probability of type1 error is called size of the test, it is denoted by and defined as
Note
is the probability of correct decision.
Type2 error
The decision relating to nonrejection of null hypo. When it is false is called type1 error.
It is denoted by and defined as
Decision  true  true 
Reject  Type1 error  Correct decision 
Do not reject  Correct decision  Type2 error 
Q18) A company of pens claims that a certain pen manufactured by him has a mean writinglife at least 460 A4 size pages. A purchasing agent selects a sample of 100 pens and put them on the test. The mean writinglife of the sample found 453 A4 size pages with standard deviation 25 A4 size pages. Should the purchasing agent reject the manufacturer’s claim at 1% level of significance?
A18)
It is given that
Specified value of population mean = = 460,
Sample size = 100
Sample mean = 453
Sample standard deviation = S = 25
The null and alternative hypothesis will be
Also, the alternative hypothesis lefttailed so that the test is left tailed test.
Here, we want to test the hypothesis regarding population mean when population SD is unknown. So we should used ttest for if writinglife of pen follows normal distribution. But it is not the case. Since sample size is n = 100 (n > 30) large so we go for Ztest. The test statistic of Ztest is given by
We get the critical value of left tailed Z test at 1% level of significance is
Since calculated value of test statistic Z (= ‒2.8) is less than the critical value
(= −2.33), that means calculated value of test statistic Z lies in rejection region so we reject the null hypothesis. Since the null hypothesis is the claim so we reject the manufacturer’s claim at 1% level of significance.
Q19) Eleven students were given a test in statistics. They were given a month’s further tuition and the second test of equal difficulty was held at the end of this. Do the marks give evidence that the students have benefitted by extra coaching?
Boys  1  2  3  4  5  6  7  8  9  10  11 
Marks I test  23  20  19  21  18  20  18  17  23  16  19 
Marks II test  24  19  22  18  20  22  20  20  23  20  17 
A19)
We compute the mean and the S.D. Of the difference between the marks of the two tests as under:
Assuming that the students have not been benefitted by extra coaching, it implies that the mean of the difference between the marks of the two tests is zero i.e.
Then, nearly and df v=111=10
Students  
1  23  24  1  0  0 
2  20  19  1  2  4 
3  19  22  3  2  4 
4  21  18  3  4  16 
5  18  20  2  1  1 
6  20  22  2  1  1 
7  18  20  2  1  1 
8  17  20  3  2  4 
9  23  23    1  1 
10  16  20  4  3  9 
11  19  17  2  3  9 




We find that (for v=10) =2.228. As the calculated value of , the value of t is not significant at 5% level of significance i.e., the test provides no evidence that the students have benefitted by extra coaching.
Q20) A college conducts both face to face and distance mode classes for a particular course indented both to be identical. A sample of 50 students of facetoface mode yields examination results mean and SD respectively as
And other sample of 100 distancemode students yields mean and SD of their examination results in the same course respectively as:
Are both educational methods statistically equal at 5% level?
A20)
Here we have
Here we wish to test that both educational methods are statistically equal. If denote the average marks of face to face and distance mode students respectively then our claim is and its complement is ≠ . Since the claim contains the equality sign so we can take the claim as the null hypothesis and complement as the alternative hypothesis. Thus,
Since the alternative hypothesis is twotailed so the test is twotailed test.
We want to test the null hypothesis regarding two population means when standard deviations of both populations are unknown. So we should go for ttest if population of difference is known to be normal. But it is not the case.
Since sample sizes are large (n1, and n2 > 30) so we go for Ztest.
For testing the null hypothesis, the test statistic Z is given by
The critical (tabulated) values for twotailed test at 5% level of significance are
Since calculated value of Z (= 2.23) is greater than the critical values
(= ±1.96), that means it lies in rejection region so we
Reject the null hypothesis i.e., we reject the claim at 5% level of significance.