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kondor19780726 [428]
3 years ago
8

Question a) Compare the Variability of Expenditure of Families in Two Towns given as follows: Number of families Expenditure (Ru

pees) 21-30 31-40 41-50 51-60 61-70 71-80 81-90 Town A 3 61 132 153 140 51 2 Town B 2 14 20 27 28 7 2
Mathematics
1 answer:
Karolina [17]3 years ago
8 0

Answer:

Kindly check explanation

Step-by-step explanation:

Given the data:

Number of families __________Expenditure

______midpoint (x) ____Town A __Town B

21-30 __25.5____________3 _______2

31-40 __35.5____________61 ______14

41-50 __45.5___________132 _____ 20

51-60 __55.5__________ 153 ______27

61-70 __65.5__________ 140 ______ 28

71-80 __75.5___________ 51 _______ 7

81-90 __85.5___________ 2 _______ 2

The variability is the variance and standard deviation :

The variance (s):

√Σ(x - m)²/Σf(x)

Town A:

m = Σ(X * f(x)) / Σf ; Σf = 542

Σ x * f(x) = (25.5*3) + (35.5*61) + (45.5*132) + (55.5*153) + (65.5*140) + (75.5*51) + (85.5*2)

= 29931 / 542

= 55. 22

Σf(x - m)² / Σf - 1 = [3(25.5-55.22)^2 + 61(35.5-55.22)^2 + 132(45.5-55.22)^2 + 153(55.5-55.22)^2 + 140(65.5-55.22)^2 + 51(75.5-55.22)^2 + 2(85.5-55.22)^2] / 541

= 76458. 4928 / 541

Variance = 141.32808

Standard deviation = √variance

Standard deviation = √141.32808

Standard deviation = 11.89

TOWN B:

m = Σ(X * f(x)) / Σf ; Σf = 100

Σ x * f(x) = (25.5*2) + (35.5*14) + (45.5*20) + (55.5*27) + (65.5*28) + (75.5*7) + (85.5*2)

= 5490 / 100

= 54.90

Variance :

Σf(x - m)² / Σf - 1 = [2(25.5-54.90)^2 + 14(35.5-54.90)^2 + 20(45.5-54.90)^2 + 27(55.5-54.90)^2 + 28(65.5-54.90)^2 + 7(75.5-54.90)^2 + 2(85.5-54.90)^2] / 99

= 16764 / 99

= 169.33

Standard deviation = √variance

Standard deviation = √169.3333

Standard deviation = 13.013

From the result above, Town A has lesser variability than Town B due to the slightly lower variance and standard deviation values.

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Test Statistics =   \frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } } follows t_n_-  _1 .

Step-by-step explanation:

We are provided with the sample data showing the math and writing scores for a sample of twelve students who took the SAT ;

Let A = Math Scores ,B = Writing Scores  and D = difference between both

So, \mu_A = Population mean for the math scores

       \mu_B = Population mean for the writing scores

 Let \mu_D = Difference between the population mean for the math scores and the population mean for the writing scores.

            <em>  Null Hypothesis, </em>H_0<em> : </em>\mu_A = \mu_B<em>     or   </em>\mu_D<em> = 0 </em>

<em>      Alternate Hypothesis, </em>H_1<em> : </em>\mu_A \neq  \mu_B<em>      or   </em>\mu_D \neq<em> 0</em>

Hence, Test Statistics used here will be;

            \frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } } follows t_n_-  _1    where, Dbar = Bbar - Abar

                                                               s_D = \sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}

                                                               n = 12

Student        Math scores (A)          Writing scores (B)         D = B - A

     1                      540                            474                                   -66

     2                      432                           380                                    -52  

     3                      528                           463                                    -65

     4                       574                          612                                      38

     5                       448                          420                                    -28

     6                       502                          526                                    24

     7                       480                           430                                     -50

     8                       499                           459                                   -40

     9                       610                            615                                       5

     10                      572                           541                                      -31

     11                       390                           335                                     -55

     12                      593                           613                                       20  

Now Dbar = Bbar - Abar = 489 - 514 = -25

 Bbar = \frac{\sum B_i}{n} = \frac{474+380+463+612+420+526+430+459+615+541+335+613}{12}  = 489

 Abar =  \frac{\sum A_i}{n} = \frac{540+432+528+574+448+502+480+499+610+572+390+593}{12} = 514

 ∑D_i^{2} = 22600     and  s_D = \sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}} = \sqrt{\frac{22600 - 12*(-25)^{2} }{12-1} } = 37.05

So, Test statistics =   \frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } } follows t_n_-  _1

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Therefore, we conclude that there is a difference between the population mean for the math scores and the population mean for the writing scores.

8 0
3 years ago
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