Answered: A. Determine the test statistic B.…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

100%

A. Determine the test statistic

B. What are the critical values

C. What is the p value

Transcribed Image Text:An evaluation was recently performed on brands and data were collected that classified each brand as being in the technology or financial institutions sector and also reported the brand value. The results in terms of value (in millions of dollars) are shown in the accompanying data table. Complete parts (a) through (c). Click the icon to view the data table. a. Assuming the population variances are equal, is there evidence that the mean brand value is different for the technology sector than for the financial institutions sector? (Use α = 0.1.) Determine the hypotheses. Let μ, be the mean brand value for the technology sector and μ₂ be the mean brand value for the financial institutions sector. Choose the correct answer below. A. Ho: H₁ = H₂ H₁: H₁ H₂ OC. Ho: H₁ H₂ H₁: H₁ H₂ Find the test statistic. tSTAT = (Round to two decimal places as needed.) Brand Values (…) 469 469 456 552 617 622 O B. Ho: Hi#H2 H₁:1₁=H₂ O D. Ho: ₁2H₂ H₁: H₁ H₂ Technology Financial Institutions 292 352 592 814 837 853 934 979 1002 1068 Print Done - X

Expert Solution

Step 1: we take the given data first then using the problem statement we define the test statistics

Given data is,

Technology	Financial Institutions
292	592
352	814
469	837
469	853
456	934
552	979
617	1002
622	1068

(a)

Let $mu subscript 1$ be the mean brand value for the technology sector

$mu subscript 2$ be the mean brand value for the financial institusions sector

Hence the null and alternative hypothesis is,

$H subscript 0 colon mu subscript 1 equals mu subscript 2 H subscript 1 colon mu subscript 1 not equal to mu subscript 2$

first we have to calculate mean and sample standard deviation for the technology sector and financial institusions sector

let,

$x subscript 1 colon t e c h n o l g y o space s e c t o r x subscript 2 colon f i n a n c i a l space i n s t i t u s i o n s space s e c t o r$

$x with bar on top subscript 1 space equals fraction numerator sum for blank of x subscript i over denominator n end fraction equals fraction numerator 292 plus 352 plus....... plus 622 over denominator 8 end fraction equals 3829 over 8 equals 478.625$

$s subscript 1 equals square root of fraction numerator 1 over denominator n minus 1 end fraction open parentheses sum for blank of x subscript i superscript 2 minus 1 over n open parentheses sum for blank of x subscript i close parentheses squared close parentheses end root space space space space equals square root of fraction numerator 1 over denominator 8 minus 1 end fraction open parentheses open parentheses left parenthesis 292 right parenthesis squared plus left parenthesis 352 right parenthesis squared plus.... plus left parenthesis 622 right parenthesis squared close parentheses minus 1 over 8 open parentheses 3829 close parentheses squared close parentheses end root space space space space equals square root of 1 over 7 open parentheses 1929303 minus 1832655.13 close parentheses end root space space space space equals square root of 1 over 7 open parentheses 96647.875 close parentheses end root space space space space equals square root of 13806.8393 end root s subscript 1 equals 117.502508$

$x with bar on top subscript 2 space equals fraction numerator sum for blank of x subscript i over denominator n end fraction equals fraction numerator 592 plus 814 plus....... plus 1068 over denominator 8 end fraction equals 7079 over 8 equals 884.875$

$s subscript 2 equals square root of fraction numerator 1 over denominator n minus 1 end fraction open parentheses sum for blank of x subscript i superscript 2 minus 1 over n open parentheses sum for blank of x subscript i close parentheses squared close parentheses end root space space space space equals square root of fraction numerator 1 over denominator 8 minus 1 end fraction open parentheses open parentheses left parenthesis 592 right parenthesis squared plus left parenthesis 814 right parenthesis squared plus.... plus left parenthesis 1068 right parenthesis squared close parentheses minus 1 over 8 open parentheses 7079 close parentheses squared close parentheses end root space space space space equals square root of 1 over 7 open parentheses 6416663 minus 6264030.13 close parentheses end root space space space space equals square root of 1 over 7 open parentheses 152632.875 close parentheses end root space space space space equals square root of 21804.6964 end root s subscript 2 equals 147.664134$

Since here assuming the population variances are equal, the test statistics is,

$t subscript S T A T end subscript equals fraction numerator x with bar on top subscript 1 minus x with bar on top subscript 2 over denominator square root of fraction numerator left parenthesis n subscript 1 minus 1 right parenthesis s subscript 1 superscript 2 plus left parenthesis n subscript 2 minus 1 right parenthesis s subscript 2 superscript 2 over denominator left parenthesis left parenthesis n subscript 1 plus n subscript 2 minus 2 right parenthesis end fraction open parentheses 1 over n subscript 1 plus 1 over n subscript 2 close parentheses end root end fraction space space space space space space space space space space equals fraction numerator 478.625 minus 884.875 over denominator square root of fraction numerator left parenthesis 8 minus 1 right parenthesis space left parenthesis 117.502508 right parenthesis squared plus left parenthesis 8 minus 1 right parenthesis space left parenthesis 147.664134 right parenthesis squared over denominator left parenthesis 8 plus 8 minus 2 right parenthesis end fraction open parentheses 1 over 8 plus 1 over 8 close parentheses end root end fraction space space space space space space space space space space space equals fraction numerator negative 406.25 over denominator square root of fraction numerator left parenthesis 7 right parenthesis space left parenthesis 13806.83929 right parenthesis plus left parenthesis 7 right parenthesis space left parenthesis 21804.69643 right parenthesis over denominator left parenthesis 14 right parenthesis end fraction open parentheses 1 fourth close parentheses end root end fraction space space space space space space space space equals fraction numerator negative 406.25 over denominator square root of fraction numerator left parenthesis 7 right parenthesis space left parenthesis 13806.83929 right parenthesis plus left parenthesis 7 right parenthesis space left parenthesis 21804.69643 right parenthesis over denominator left parenthesis 14 right parenthesis end fraction open parentheses 1 fourth close parentheses end root end fraction space space space space space space space space space space space equals fraction numerator negative 406.25 over denominator square root of fraction numerator left parenthesis 249280.75 right parenthesis space over denominator left parenthesis 14 right parenthesis end fraction open parentheses 1 fourth close parentheses end root end fraction space space space space space space space space space space equals fraction numerator negative 406.25 over denominator square root of fraction numerator 17805.76786 over denominator 4 end fraction end root end fraction space space space space space space space space space space equals fraction numerator negative 406.25 over denominator 66.71912742 end fraction t subscript S T A T end subscript equals negative 6.088958529 almost equal to negative 6.09 t subscript S T A T end subscript equals negative 6.09$

Hence the test statistics value is,

$bold italic t subscript bold S bold T bold A bold T end subscript bold equals bold minus bold 6 bold. bold 09$

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