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Dear Sir,

Please provide answers to the below set of questions.

1. A sum of `8550 is to be paid in 15 installments where each installment is `10 more than the previous installment. Find the first installment and the last installment.

2. A salesman is known to sell a product in 3 out of 5 attempts. While another salesman in 2 out of 5 attempts. Find the probability that

a. No sales will happen

b. Either of them will succeed in selling the product

3. A hundred squash balls are tested by dropping from a height of 100 inches and measuring the height of the bounce. A ball is “fast” if it rises above 32 inches. The average height of bounce was 30 inches and the standard deviation was ¾ inches. What is the chance of getting a “fast” standard ball?

4. Explain the chi-square testing- (i) as a test for independence of attributes, and (ii) as a test for goodness of fit.

regards,

Mythri

Solution

A sum of `8550 is to be paid in 15 installments where each installment is `10 more than the previous installment. Find the first installment and the last installment.

Solution: Since the each installment is increases by 10 than the previous installment,

Therefore, d =10

Let the first installment is = a

No. of installments is, n = 15

This forms an A.P. Then we have

S_n= n/2 { 2a+(n-1)d}

8550= 15/2 { 2a+(15-1)10}

8550= 15/2 { 2a+14*10}

8550*2= 15 { 2a+140}

17100=30a+2100

17100-2100=30a

a=500

Since the first Installment is a =500

So the last installment will be

=a+(a+d)+ (a+2d)+(a+3d)+(a+4d)+(a+5d)+(a+6d)+⋯……………..+(a+14d)

last instalment=(a+14d)

=(500+14*10)

=640

First installment = Rs. 500

Last installment = Rs. 640

A salesman is known to sell a product in 3 out of 5 attempts. While another salesman in 2 out of 5 attempts. Find the probability that

No sales will happen

Solution: Let,

A represent the event that first salesman is able to sell the product

A^c Represent the event that first salesman is not able to sell the product

B represent the event that second salesman is able to sell the product

B^c Represent the event that second salesman is not able to sell the product

It is given that

P(A)=3/5 and P(B)=2/5

Therefore,

P(A^c )=1- 3/5=2/5 and P(B^c )=1-2/5=3/5

The required probabilities are calculated as follows:

P ( no sale will be affected when they both try to sell the product)

= P(A^c B^c )= P(A^c ) P (B^c) (By multiplication theorem)

=2/5*3/5

=6/25

= 0.24

Either of them will succeed in selling the product

P (either of them will succeed in selling the product)

=P(A)+ P(B)- P(AB) (by theorem of total probability)

=3/5+2/5- P(A) P(B) (by multiplication theorem)

=1-(3/5)(2/5)

=1-6/25

=19/25

=0.76

A hundred squash balls are tested by dropping from a height of 100 inches and measuring the height of the bounce. A ball is “fast” if it rises above 32 inches. The average height of bounce was 30 inches and the standard deviation was ¾ inches. What is the chance of getting a “fast” standard ball?

Solution: Let x be the normal variable with mean μ and Standard Deviation,σ.

Therefore, x ~N (μ,σ^2)

Here, total no. of observations N = 100

Mean, μ=30 inches ; Standard Deviation,σ=3/4 inches=0.75 inches

Also, we will convert x into standard normal variable Z such that

Z= ((x-μ))/σ

We shall find the no. of observations likely to exceed 32 inches, i.e. ball is known as “ fast standard ball”.

P (x>32)= P [((x-μ))/σ> (32-30)/0.75 ]

=P (Z>2.7)

=0.5-F(2.7)

from the table,we got

=0.5-0.4808

=0.0192

Therefore,the chance of getting the fast standard ball i.e. if a ball rise above 32 inches.

=N*P (x>32)z

=100*0.0192

=1.92

≅2 observations

Explain the chi-square testing- (i) as a test for independence of attributes, and (ii) as a test for goodness of fit.

Solution: Chi-Square Testing

As a test for independence of attributes

If there are two categorical variables, and our interest is to examine whether these two variables are associated with each other, the chi-square (χ^2 ) test of independence is the correct tool to use. This test is very popular in analyzing cross-tabulations in which an investigator is keen to find out whether the two attributes of interest have any relationship with each other.

The cross-tabulation is popularly called by the term “contingency table”. It contains frequency data that correspond to the categorical variables in the row and column. The marginal totals of the rows and columns are used to calculate the expected frequencies that will be part of the computation of the c² statistic. For calculations on expected frequencies, refer hyper stat on χ^2 test.

Example: A marketing firm producing detergents is interested in studying the consumer behavior in the context of purchase decision of detergents in a specific market. This company is a major player in the detergent market that is characterized by intense competition. It would like to know in particular whether the income level of the consumers influence their choice of the brand. Currently there are four brands in the market. Brand 1 and Brand 2 are the premium brands while Brand 3 and Brand 4 are the economy brands.

A representative stratified random sampling procedure was adopted covering the entire market using income as the basis of selection. The categories that were used in classifying income level are: Lower, Middle, Upper Middle and High. A sample of 600 consumers participated in this study. The following data emerged from the study.

Cross Tabulation of Income versus Brand chosen (Figures in the cells represent number of consumers)

Brands

Brand 1 Brand 2 Brand 3 Brand 4 Total

Income

Lower 25 15 55 65 160

Middle 30 25 35 30 120

Upper Middle 50 55 20 22 147

Upper 60 80 15 18 173

Total 165 175 125 135 600

Analyze the cross-tabulation data above using chi-square test of independence and draw your conclusions.

Solution:

Null Hypothesis: There is no association between the brand preference and income level (These two attributes are independent).

Alternative Hypothesis: There is association between brand preference and income level (These two attributes are dependent).

Let us take a level of significance of 5%.

In order to calculate the χ^2 value, you need to work out the expected frequency in each cell in the contingency table. In our example, there are 4 rows and 4 columns amounting to 16 elements. There will be 16 expected frequencies. For calculating expected frequencies, please go through hyper stat. Relevant data tables are given below:

Observed Frequencies (These are actual frequencies observed in the survey)

Brands

Brand 1 Brand 2 Brand 3 Brand 4 Total

Income

Lower 25 15 55 65 160

Middle 30 25 35 30 120

Upper Middle 50 55 20 22 147

Upper 60 80 15 18 173

Total 165 175 125 135 600

Expected Frequencies (These are calculated on the assumption of the null hypothesis being true: That is, income level and brand preference are independent)

Brands

Brand 1 Brand 2 Brand 3 Brand 4 Total

Income

Lower 44.000 46.667 33.333 36.000 160.000

Middle 33.000 35.000 25.000 27.000 120.000

Upper Middle 40.425 42.875 30.625 33.075 147.000

Upper 47.575 50.458 36.042 38.925 173.000

Total 165.000 175.000 125.000 135.000 600.000

Calculation: Compute

x^2=Σ(〖(O-E)〗^2/E

There are 16 observed frequencies (O) and 16 expected frequencies (E). As in the case of the goodness of fit, calculate this χ^2 value. In our case, the computed χ^2 =131.76 as shown below: Each cell in the table below shows (O-E) ²/ (E)

Brand 1 Brand 2 Brand 3 Brand 4

Income

Lower 8.20 21.49 14.08 23.36

Middle 0.27 2.86 4.00 0.33

Upper Middle 2.27 3.43 3.69 3.71

Upper 3.24 17.30 12.28 11.25

And there are 16 such cells. Adding all these 16 values, we get χ^2 =131.76

The critical value of χ^2depends on the degrees of freedom. The degrees of freedom = (the number of rows-1) multiplied by (the number of colums-1) in any contingency table. In our case, there are 4 rows and 4 columns. So the degrees of freedom =(4-1). (4-1) =9. At 5% level of significance, critical χ^2 for 9 d.f = 16.92. Therefore reject the null hypothesis and accept the alternative hypothesis.

The inference is that brand preference is highly associated with income level. Thus, the choice of the brand depends on the income strata. Consumers in different income strata prefer different brands. Specifically, consumers in upper middle and upper income group prefer premium brands while consumers in lower income and middle-income category prefer economy brands. The company should develop suitable strategies to position its detergent products. In the marketplace, it should position economy brands to lower and middle-income category and premium brands to upper middle and upper income category.

As a test for goodness of fit

The Chi-square test used with one sample is described as a "goodness of fit" test. It can help you decide whether a distribution of frequencies for a variable in a sample is representative of, or "fits", a specified population distribution. A number of marketing problems involve decision situations in which it is important for a marketing manager to know whether the pattern of frequencies that are observed fit well with the expected ones. The appropriate test is the χ^2 test of goodness of fit. The illustration given below will clarify the role of χ^2 in which only one categorical variable is involved.

Example: In consumer marketing, a common problem that any marketing manager faces is the selection of appropriate colors for package design. Assume that a marketing manager wishes to compare five different colors of package design. He is interested in knowing which of the five the most preferred one is so that it can be introduced in the market. A random sample of 400 consumers reveals the following:

Package color Preference by customers

Red 70

Blue 106

Green 80

Pink 70

Orange 74

Total 400

Do the consumer preferences for package colors show any significant difference?

Solution: If you look at the data, you may be tempted to infer that Blue is the most preferred color. Statistically, you have to find out whether this preference could have arisen due to chance. The appropriate test statistic is the χ^2 test of goodness of fit.

Null Hypothesis: All colors are equally preferred.

Alternative Hypothesis: They are not equally preferred

Package color Observed Frequencies (O) Expected Frequencies (E) (〖O-E)〗^2 x^2=Σ(〖(O-E)〗^2/E

Red 70 80 100 1.250

Blue 106 80 676 8.450

Green 80 80 0 0.000

Pink 70 80 100 1.250

Orange 74 80 36 0.450

Total 400 400 11.400

Please note that under the null hypothesis of equal preference for all colors being true, the expected frequencies for all the colors will be equal to 80. Applying the formula

x^2=Σ(〖(O-E)〗^2/E

We get the computed value of chi-square χ^2 = 11.400

The critical value of χ^2 at 5% level of significance for 4 degrees of freedom is 9.488. So, the null hypothesis is rejected. The inference is that all colors are not equally preferred by the consumers. In particular, Blue is the most preferred one. The marketing manager can introduce blue color package in the market.

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