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A survey was conducted two years ago asking college students their top motivations for using a credit card. To determine whet

critical value is 7.779 I think

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Answer #1

Computational Table:

Response Old Survey % Frequency (Oi) Expected Frequency (Ei) Oi - Ei (Oi-Ei)2 (Oi-Ei)2/Ei
Rewards 27% 111 27% of 425 = 114.25 -3.75 14.06 0.12
Low rates 24% 96 24% of 425 = 102 -6 36.00 0.35
Cash back 22% 107 22% of 425 = 93.5 13.5 182.25 1.95
Discounts 7% 47 7% of 425 = 29.75 17.25 297.56 10.00
Other 20% 64 20% of 425 = 85 -21 441.00 5.19
Total 100% 425 425 17.615

Test statistic:

\chi ^{2} = \frac{\sum (Oi-Ei)^{2}}{Ei} = 17.615

Degrees of Freedom: k-1 = 5-1 = 4

Where, K = Number of Responses = 5

Critical value: For \alpha = 0.10

\chi _{\alpha ,df}^{2} = \chi _{0.10 ,4}^{2} = 7.779 ..........Using Chi square table

Conclusion:

Test statistic > critical value, i.e 17.615 > 7.779, That is Reject Ho at 10% level of significance.

Reject Ho. At the 10% level of significance level, there is enough evidence to conclude that the distribution of motivations differs from the claimed or expected distribution.

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