plese answer all 8 questions do not just answer 1. thank you. 1. Solution Null h

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plese answer all 8 questions do not just answer 1. thank you.

1.

Explanation / Answer

Null hypothesis: In Chi-Square goodness of fit test, the null hypothesis assumes that there is no significant difference between the observed and the expected value. Alternative hypothesis: In Chi-Square goodness of fit test, the alternative hypothesis assumes that there is a significant difference between the observed and the expected value. Sno Believe percentage truck 28% car 54% other 18% And a random sample of 600 were taken and results are as follows. Sno Observed Frequency truck 190 car 292 other 118 Now, expected frequency can be obtained if we multiply believe percentage with size of sample taken that is 600. Sno Expected Frequency truck 168 car 324 other 108 Answer 1: As we can see from the expected frequency table that the expected count for those who use car as primary   transportation are 324 people. Answer 2: The value of chi-square statistic is given as follows: Sno Observed Frequency(Oi) Expected Frequency(Ei) (Oi-Ei)2/Ei truck 190 168 2.88 car 292 324 3.16 other 118 108 0.93 Chi-Square Statistic = Total Sum -> 6.97 Chi-Square statistic = 6.97 Answer 3: As we have 3 observations that is truck, car and other and degree of freedom is given by n-1 = 3-1 = 2 Therefore, degree of freedom = 2 Answer 4: We can calculate the critical value in MS-Excel by using the function "CHIINV()" which takes (? = 0.05, degree of freedom = 2) as inputs and outputs the critical value. Critical value = CHIINV(0.05,2) = 5.9915 We observed that Chi-Square value is greater than critical value, so we reject our null hypothesis and we conclude that there is a significant difference between the observed and the expected value.

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