Unit – 5

Applied Statistics

5.1 Curve Fitting by the Method of Least Squares- Fitting Of Straight Lines

Curve Fitting of Type y=ax^b Algorithm

In this article we are going to develop an algorithm for fitting curve of type

y = axb using least square regression method.

Procedure for fitting y = axb

We have,

Taking log on both side of equation (1), we get

5.2 Second Degree Parabolas and More General Curves

Second-order curve

A plane curve whose rectangular Cartesian coordinates fulfil an algebraic calculation of the second degree:

Equation need not define a real geometrical form, but to reserve generality in such situations one says that it describes an imaginary second-order curve. Depending on the values of the coefficients of the equation  it can be changed by parallel movement and rotation of the coordinate system through some angle to one of the 9 canonical forms given under, to each of which there agrees a definite class of curves. Viz.,

Non-degenerate curves:

, ellipses (cf. Ellipse);

, hyperbolas (cf. Hyperbola);

, parabolas (cf. Parabola);

, imaginary ellipses;

Degenerate curves:

, pairs of imaginary intersecting lines;

, pairs of real intersecting lines;

, pairs of real parallel lines;

, pairs of imaginary parallel lines;

,a pair of coincident real lines.

A second-order curve that has a unique centre of symmetry (the centre of the second-order curve) is named a central curve. The coordinates of the centre of a second-order curve are determined by the explanation of the system

A second-order curve without a centre of symmetry or with an indefinite centre is named a non-central curve.

Soundings on the form of a second-order curve can be approved out without reducing the general equation to canonical form. This is done by seeing the values of the so-called fundamental invariants of a second-order curve, that is, the resulting expressions in the coefficients of the equation , the values of which are not changed under a parallel displacement and a rotation of the coordinate system :

And the semi-invariant

Which is an invariant with respect to rotation of the coordinate system (see the Table below).

Many significantbelongings of second-order curves can be calculated by means of the characteristic quadratic form

Corresponding to equation . In specific, a non-degenerate second-order curve turns out to be an ellipse, an imaginary ellipse, a hyperbola, or a parabola, dependent on whether   is a positive-definite, a negative-definite, an indefinite, or a semi-definite quadratic form. This can be recognized from the roots of its characteristic equation:

=0 or 

The three fundamental invariants ,  and  define a second-order curve (without the case of parallel lines) up to a motion of the Euclidean plane: If the agreeing invariants ,  and  of two curves are equal, then these curves can be made to coincide by a motion. In other words, these curves are corresponding with respect to the group of motions of the plane (metrically equivalent).

There is a ordering of second-order curves from the point of view of other groups of alterations. Thus, with respect to the more general (than the group of motions) group of affine alterations any two curves distinct by equations of the same canonical form are correspondent. For example, two similar second-order curves are observed as equivalent. The contacts between different affine classes of second-order curves make it likely to formarrangement from the point of view of projective geometry, in which elements at infinity do not play a singular role. The real non-degenerate second-order curves, ellipses, hyperbolas, and parabolas, form one projective class — the class of real oval curves (ovals, cf. Oval). A real oval curve is an ellipse, hyperbola or parabola, dependent on how it is placed with respect to the line at infinity; an ellipse intersects the line at infinity in two imaginary points, a hyperbola intersects it in two real points and a parabola touches it; there are projective alterations that take these curves into one another. There are 5 projective equivalence classes of second-order curves. Viz.,

Non-degenerate curves ( are the homogeneous coordinates):

, a real oval;

, an imaginary oval;

Degenerate curves:

, a pair of real lines;

, a pair of imaginary lines;

, a pair of coincident lines.

Apart from the analytic method of defining second-order curves (specifying the equation) there are other methods. For example, an ellipse, hyperbola and parabola can be obtained as a section of a conical surface by a plane (see Conic sections).'

Non-degenerate curves  Degenerate curves  Central curves ( )  Real ellipses Pairs of intersecting straight lines (a real point)      Imaginary ellipses     Hyperbolas Pairs of real non-intersecting straight lines Non-central curves ( ) Parabolas  Pairs of imaginary parallel lines        Pairs of real parallel lines        Pairs of coincident parallel lines

5.3 Test of Significance: Large Sample Test for Single Proportion

Large Sample Test for a Proportion

• The sampling distribution of  is approximately normal.
• Use the value of p in the null hypothesis when computing the standard deviation of .
• The test statistic is

Hypothesis Test: Difference Between Proportions

This lesson describes how to conduct a hypothesis test to define whether the difference between two proportions is significant.

The test procedure, called the two-proportion z-test, is suitable when the following conditions are met:

• The sampling method for each population is simple random sample.
• The examples are independent.
• Each example comprises at least 10 attainments and 10 failures.
• Each population is at least 20 times as big as its example.

This approach contains of four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results.

Test Your Considerate

In this section, two sample difficulties illustrate how to conduct a hypothesis test for the difference between two sizes. The first problem includes two-tailed test; the second problem, a one-tailed test.

Problem 1: Two-Tailed Test

Suppose Acme Drug Company is making a new medicine designed to prevent colds. The company claims that the drug is equally effective for men and women. To test this claim, they chose a simple random sample of 100 women and 200 men from a population of 100,000 volunteers.

At the end of the study, 38% of the women had a cold; and 51% of men caught a cold. Based on these results, can we rule out the company's claim that the drug is equally effective for men and women? Use a significance level of 0.05.

Solution: The solution to this problem involves four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze the sample data, and (4) interpret the results. We work through the following steps:

 State the hypotheses. The first step is to affirm the null hypothesis and an alternative hypothesis.

Null hypothesis: P1 = P2

Alternative hypothesis: P1 ≠ P2

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion of population 1 is too large or too small.

 Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-sided z-test.

 Analyze the sample data. Using the sample data, we calculated the proportion of the pooled sample (p) and the standard error (SE). Using these measures, we compute the z-score (z) test statistic.

p = (p1 * n1 + p2 * n2) / (n1 + n2)

p = [(0.38 * 100) + (0.51 * 200)] / (100 + 200)

p = 140/300 = 0.467

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

SE = sqrt[ 0.467 * 0.533 * ( 1/100 + 1/200 ) ]

SE = sqrt [0.003733] = 0.061

z = (p1 - p2) / SE = (0.38 - 0.51)/0.061 = -2.13

Suppose Acme Drug Company is making a new medicine designed to prevent colds. The company claims that the drug is equally effective for men and women. To test this claim, they chose a simple random sample of 100 women and 200 men from a population of 100,000 volunteers. At the end of the study, 38% of the women had a cold; and 51% of men caught a cold. Based on these results, can we rule out the company's claim that the drug is equally effective for men and women? Use a significance level of 0.05. Solution: The solution to this problem involves four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze the sample data, and (4) interpret the results. We work through the following steps:

 State the hypotheses. The first step is to affirm the null hypothesis and an alternative hypothesis. Null hypothesis: P1 = P2 Alternative hypothesis: P1 ≠ P2 Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion of population 1 is too large or too small.

 Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-sided z-test.

Analyze the sample data. Using the sample data, we calculated the proportion of the pooled sample (p) and the standard error (SE). Using these measures, we compute the z-score (z) test statistic. The analysis described above is a two-proportion z-test.

Interpretation of results If the sample results are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. Typically, this involves comparing the P value with the significance level and rejecting the null hypothesis when the P value is less than the significance level. Hypothesis test: difference between proportions This lesson explains how to perform a hypothesis test to determine if the difference between two proportions is significant. The test procedure, called the two-sided z-test, is appropriate when the following conditions are met:

 The sampling method for each population is a simple random sampling.

 The samples are independent.

 Each champion includes at least 10 successes and 10 failures.

 Each population is at least 20 times larger than its sample.

This approach involves four phases: (1) declaring the hypotheses, (2) formulating an analysis plan, (3) analyzing the sample data, and (4) interpreting the results. State hypotheses Each hypothesis test requires the analyst to declare a null hypothesis and an alternative hypothesis. The following table shows three sets of hypotheses. Each makes a statement about the difference d between two proportions of the population, P1 and P2. (In the table, the symbol ≠ means "not equal to").

• The first set of hypotheses (Set 1) is an example of a two-tailed test, since an extreme value on both sides of the sampling distribution would induce an investigator to reject the null hypothesis. The other two sets of hypotheses (sets 2 and 3) are one-tailed tests, since an extreme value on one side of the sampling distribution would induce an investigator to reject the null hypothesis.
• When the null hypothesis states that there is no difference between the two proportions of the population (i.e., d = P1 - P2 = 0), the null and alternative hypothesis for a two-tailed test is often stated as follows.
• I have: P1 = P2
• Has: P1 ≠ P2
• Formulate an analysis plan.
• The analysis plan describes how to use the sample data to accept or reject the null hypothesis. You must specify the following elements.
•  Level of significance. Researchers often choose significance levels of 0.01, 0.05, or 0.10; but you can use any value between 0 and 1.
•  Test method. Use the two aspect ratio z test (described in the next section) to determine if the assumed difference between population proportions differs significantly from the observed difference in the sample.
• Analyze the sample data.
• Using the sample data, complete the following calculations to find the test statistic and associated P-value.
•  Proportion of the aggregate sample. Since the null hypothesis states that P1 = P2, we use an aggregate sample proportion (p) to calculate the standard error of the sampling distribution.
• p = (p1 * n1 + p2 * n2) / (n1 + n2)
• Where p1 is the proportion of the sample from population 1, p2 is the proportion of the sample from population 2, n1 is the size of sample 1 and n2 is the size of sample 2.
•  Standard error. Calculate the standard error (SE) of the sampling distribution difference between two proportions.
• SE = sqrt {p * (1 - p) * [(1 / n1) + (1 / n2)]}
• Where p is the proportion of the pooled sample, n1 is the size of sample 1 and n2 is the size of sample 2.
•  Statistical test. The test statistic is a z (z) score defined by the following equation.
• z = (p1 - p2) / SE
• Where p1 is the proportion of sample 1, p2 is the proportion of sample 2 and SE is the standard error of the sampling distribution.
• p-value The P value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a z-score, use the normal distribution calculator to assess the probability associated with the z-score. (See the example problems at the end of this lesson for examples of how this is done.)
• The analysis described above is a two-way z-test.
• Interpretation of results
• If the sample results are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. Typically, this involves comparing the P value with the significance level and rejecting the null hypothesis when the P value is less than the significance level.
• Test your understanding
• In this section, two sample problems illustrate how to perform a hypothesis test for the difference between two proportions. The first problem involves a two-tailed test; The second problem, a tail test.
• Problem 1: two-tailed test
• Suppose Acme Drug Company develops a new drug, designed to prevent colds. The company claims that the drug is equally effective for men and women. To test this claim, they chose a simple random sample of 100 women and 200 men from a population of 100,000 volunteers.
• At the end of the study, 38% of the women had a cold; and 51% of men caught a cold. Based on these results, can we rule out the company's claim that the drug is equally effective for men and women? Use a significance level of 0.05.
• Solution: The solution to this problem involves four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze the sample data, and (4) interpret the results. We work through the following steps:
•  State the hypotheses. The first step is to affirm the null hypothesis and an alternative hypothesis.
• Null hypothesis: P1 = P2
• Alternative hypothesis: P1 ≠ P2
• Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion of population 1 is too large or too small.
•  Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-sided z-test.
•  Analyze the sample data. Using the sample data, we calculated the proportion of the pooled sample (p) and the standard error (SE). Using these measures, we compute the z-score (z) test statistic.
• p = (p1 * n1 + p2 * n2) / (n1 + n2)
• p = [(0.38 * 100) + (0.51 * 200)] / (100 + 200)
• p = 140/300 = 0.467
• SE = sqrt {p * (1 - p) * [(1 / n1) + (1 / n2)]}
• SE = sqrt [0.467 * 0.533 * (1/100 + 1/200)]
• SE = sqrt [0.003733] = 0.061
• z = (p1 - p2) / SE = (0.38 - 0.51) /0.061 = -2.13
• Where p1 is the proportion of the sample in sample 1, where p2 is the proportion of the sample in sample 2, n1 is the size of sample 1 and n2 is the size of sample 2.
• Since we have a two-tailed test, the P value is the probability that the z-score is less than -2.13 or greater than 2.13.
• We use the normal distribution calculator to find P (z <-2.13) = 0.017 and P (z> 2.13) = 0.017. Therefore, the value P = 0.017 + 0.017 = 0.034.
•  Interpret the results. Since the P value (0.034) is lower than the significance level (0.05), we cannot accept the null hypothesis.
• Note: If you use this approach during an exam, you can also mention why this approach is appropriate. In particular, the approach is appropriate because the sampling method was simple random sampling, the samples were independent, each population was at least 10 times larger than its sample, and each sample included at least 10 successes and 10 failures.
• Problem 2: testing a queue
• Suppose the previous example is declared slightly differently. Suppose Acme Drug Company develops a new drug, designed to prevent colds. The company claims that the drug is more effective for women than for men. To test this claim, they chose a simple random sample of 100 women and 200 men from a population of 100,000 volunteers.
• At the end of the study, 38% of the women had a cold; and 51% of men caught a cold. Based on these results, can we conclude that the medication is more effective for women than for men? Use a significance level of 0.01.
• Solution: The solution to this problem involves four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze the sample data, and (4) interpret the results. We work through the following steps:
•  State the hypotheses. The first step is to affirm the null hypothesis and an alternative hypothesis.
• Null hypothesis: P1> = P2
• Alternative hypothesis: P1 <P2
• Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the percentage of women who have a cold (p1) is sufficiently lower than the proportion of men who have a cold (p2).
•  Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a two-sided z-test.
•  Analyze the sample data. Using the sample data, we calculated the proportion of the pooled sample (p) and the standard error (SE). Using these measures, we compute the z-score (z) test statistic.

p = (p1 * n1 + p2 * n2) / (n1 + n2)

p = [(0.38 * 100) + (0.51 * 200)] / (100 + 200)

p = 140/300 = 0.467

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

SE = sqrt[ 0.467 * 0.533 * ( 1/100 + 1/200 ) ]

SE = sqrt [0.003733] = 0.061

z = (p1 - p2) / SE = (0.38 - 0.51)/0.061 = -2.13

Where p1 is the proportion of the sample in sample 1, where p2 is the proportion of the sample in sample 2, n1 is the size of sample 1 and n2 is the size of sample 2.

Subsequently we have a one-tailed test, the P significance is the probability that the z-score is less than -2.13. We practice the normal distribution calculator to discover P (z <-2.13) = 0.017. Therefore, the P value = 0.017.

 Interpret the results. Since the P value (0.017) is greater than the significance level (0.01), we cannot reject the null hypothesis

5.4 Difference of Proportions

To estimate the difference between two population proportions with a confidence interval, you can use the Central Limit Theorem when the sample sizes are large enough (typically, each at least 30). When a statistical characteristic, such as opinion on an issue (support/don’t support), of the two groups being compared is categorical, people want to report on the differences between the two population proportions — for example, the difference between the proportion of women and men who support a four-day work week. How do you do this?

You estimate the difference between two population proportions, p1 – p2, by taking a sample from each population and using the difference of the two sample proportions,

Plus or minus a margin of error. The result is called a confidence interval for the difference of two population proportions, p1 – p2.

The formula for a confidence interval (CI) for the difference between two population proportions is

And n1 are the sample proportion and sample size of the first sample, and

And n2 are the sample proportion and sample size of the second sample. The value z* is the appropriate value from the standard normal distribution for your desired confidence level. (Refer to the following table for z*-values.)

z*–values for Various Confidence Levels

To calculate a CI for the difference between two population proportions, do the following:

1. Determine the confidence level and find the appropriate z*-value.

Refer to the above table.

For the first sample by taking the total number from the first sample that are in the category of interest and dividing by the sample size, n1. Similarly, find for the second sample.

And divide that by n1. Find

And divide that by n2. Add these two results together and take the square root.

This step gives you the margin of error.

Plus or minus the margin of error from Step 5 to obtain the CI.

The lower end of the CI is

Minus the margin of error, and the upper end of the CI is

Plus the margin of error.

The formula shown here for a CI for p1 – p2 is used under the condition that both of the sample sizes are large enough for the Central Limit Theorem to be applied and allow you to use a z*-value; this is true when you are estimating proportions using large scale surveys, for example. For small sample sizes, confidence intervals are beyond the scope of an intro statistics course.

Suppose you work for the Las Vegas Chamber of Commerce, and you want to estimate with 95% confidence the difference between the percentage of all females who have ever gone to see an Elvis impersonator and the percentage of all males who have ever gone to see an Elvis impersonator, in order to help determine how you should market your entertainment offerings.

1. Because you want a 95% confidence interval, your z*-value is 1.96.
2. Suppose your random sample of 100 females includes 53 females who have seen an Elvis impersonator, so

Is 53 divided by 100 = 0.53. Suppose also that your random sample of 110 males includes 37 males who have ever seen an Elvis impersonator, so

Is 37 divided by 110 = 0.34.

To interpret these results within the context of the problem, you can say with 95% confidence that a higher percentage of females than males have seen an Elvis impersonator, and the difference in these percentages is somewhere between 6% and 32%, based on your sample.

The temptation is to say, “Well, I knew a greater proportion of women has seen an Elvis impersonator because that sample proportion was 0.53 and for men it was only 0.34. Why do I even need a confidence interval?” All those two numbers tell you is something about those 210 people sampled. You also need to factor in variation using the margin of error to be able to say something about the entire populations of men and women.

For example, if you had switched the males and females, you would have gotten –0.19 for this difference. That’s okay, but you can avoid negative differences in the sample proportions by having the group with the larger sample proportion serve as the first group (here, females).

However, even if the group with the larger sample proportion serves as the first group, sometimes you will still get negative values in the confidence interval. Suppose in the above example that only 0.43 of women had seen an Elvis impersonator. Thus, the difference in proportions is 0.09, and the upper end of the confidence interval is 0.09 + 0.13 = 0.22 while the lower end is 0.09 – 0.13 = –0.04. This means that the true difference is reasonably anywhere from 22% more women to 4% more men. It’s too close to tell for sure.

5.5 Single Mean

The single mean (or one-sample) t-test is used to compare the mean of a variable in a sample of data to a (hypothesized) mean in the population from which our sample data are drawn. This is important because we seldom have access to data for an entire population. The hypothesized value in the population is specified in the Comparison value box.

We can perform either a one-sided test (i.e., less than or greater than) or a two-sided test (see the Alternative hypothesis dropdown). We use one-sided tests to evaluate if the available data provide evidence that the sample mean is larger (or smaller) than the comparison value (i.e., the population value in the null-hypothesis)

Example :

We have access to data from a random sample of grocery stores in the UK. Management will consider entering this market if consumer demand for the product category exceeds 100M units, or, approximately, 1750 units per store. The average demand per store in the sample is equal to 1953. While this number is larger than 1750 we need to determine if the difference could be attributed to sampling error.

You can find the information on unit sales in each of the sample stores in the demand_uk.rda data set. The data set contains two variables, store_id and demand_uk. Our null-hypothesis is that the average store demand in the UK is equal to 1750 unit so we enter that number into the Comparison value box. We choose the Greater than option from the Alternative hypothesis drop-down because we want to determine if the available data provides sufficient evidence to reject the null-hypothesis favor of the alternative that average store demand in the UK is larger than 1750.

The first two blocks of output show basic information about the test (e.g.,. The null and alternative hypothesis) and summary statistics (e.g., mean, standard deviation, standard error, margin or error, etc.). The final row of output shows the following:

• Diff is the difference between the sample mean (1953.393) and the comparison value (1750)
• Se is the standard error (i.e., the standard deviation of the sampling distribution of diff)
• t.value is the t statistic associated with diff that we can compare to a t-distribution (i.e., diff / se)
• p.value is the probability of finding a value as extreme or more extreme than diff if the null hypothesis is true
• Df is the degrees of freedom associated with the statistical test (i.e., n - 1)
• 5% 100% show the 95% confidence interval around the sample mean (1897 to Inf.). These numbers provide a range within which the true population mean is likely to fall

5.6 Difference of Means and Difference of Standard Deviations

Significance test of difference between sample means

Given two independent examples and with means standard derivations from a normal population with the same variance, we have to test the hypothesis that the population means   are same For this, we calculate

It can be shown that the variate t defined by (1) follows the t distribution with degrees of freedom.

If the calculated value the difference between the sample means is said to be significant at 5% level of significance.

If , the difference is said to be significant at 1% level of significance.

If the data is said to be consistent with the hypothesis that .

Cor. If the two samples are of same size and the data are paired, then t is defined by

=difference of the ith member of the sample

d=mean of the differences = and the member of d.f.=n-1.

Example 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?

Sol. 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=11-1=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

From table IV, 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.

Example:

From a random sample of 10 pigs fed on diet A, the increase in weight in certain period were 10,6,16,17,13,12,8,14,15,9 lbs. For another random sample of 12 pigs fed on diet B, the increase in the same period were 7,13,22,15,12,14,18,8,21,23,10,17 lbs. Test whether diets A and B differ significantly as regards their effect on increases in weight ?

Sol. We calculate the means and standard derivations of the samples as follows

	Diet A			Diet B
10	-2	4	7	-8	64
6	-6	36	13	-2	4
16	4	16	22	7	49
17	5	25	15	0	0
13	1	1	12	-3	9
12	0	0	14	-1	1
8	-4	16	18	3	9
14	2	4	8	-7	49
15	3	9	21	6	36
9	-3	9	23	8	64
			10	-5	25
			17	2	4
120			180	0	314

Assuming that the samples do not differ in weight so far as the two diets are concerned i.e.

For v=20, we find =2.09

The calculated value of

Hence the difference between the samples means is not significant i.e. thew two diets do not differ significantly as regards their effects on increase in weight.

References

1. E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
2. P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
3. S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
4. W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
5. N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
6. B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.
7. T. Veerarajan, “Engineering Mathematics”, Tata Mcgraw-Hill, New Delhi, 2010