Question

In: Statistics and Probability

An evaluation was recently performed on brands and data were collected that classified each brand as...

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).

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=0.05​.) Determine the hypotheses. Let μ1 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

-Find the test statistic

-Find the p-Value


b. Repeat​ (a), assuming that the population variances are not equal.


c. Compare the results of​ (a) and​ (b).

Technology Financial Institutions
255 577
359 836
417 806
476 889
492 998
564 976
634 1022
631 1074

Solutions

Expert Solution

Technology ( X )

Financial Institutions ( Y )
255 49952.25 577 102560.0625
359 14280.25 836 3751.5625
417 3782.25 806 8326.5625
476 6.25 889 68.0625
492 182.25 998 10150.5625
564 7310.25 976 6201.5625
634 24180.25 1022 15562.5625
631 23256.25 1074 31240.5625
Total 3828 122950 7178 177861.5







To Test :-

H0 :-  

H1 :-  

Test Statistic :-





t = -5.7135


Test Criteria :-
Reject null hypothesis if   


Result :- Reject Null Hypothesis

P value = 0.0000536

Decision based on P value

Reject null hypothesis if P value < α=0.05 level of significance

0.0000536 < 0.05, we reject null hypothesis

Conclusion :- Accept Alternative Hypothesis

There is sufficient evidence to support  the claim that the mean brand value is different for the technology sector than for the financial institutions​ sector.

Population variances are not equal.

Test Statistic :-


t = -5.7135


Test Criteria :-
Reject null hypothesis if   


DF = 13


Result :- Reject Null Hypothesis

P value = 0.00007131

Decision based on P value

Reject null hypothesis if P value < α=0.05 level of significance

0.00007131 < 0.05, we reject null hypothesis

Conclusion :- Accept Alternative Hypothesis

There is sufficient evidence to support  the claim that the mean brand value is different for the technology sector than for the financial institutions​ sector.

Part c)

In both the method we are reject the null hypothesis.


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