A particular city has an Asian population of 1419 people, out of a total populat

Question

A particular city has an Asian population of 1419 people, out of a total population of 23,609. Conduct a goodness of fit test at the 5% level to determine if the self-reported sub-groups of Asians are evenly distributed. Round expected frequency to two decimal places.

Part (a)

State the null hypothesis. Choose 1 or 2

1. The self-reported sub-groups of Asians are not evenly distributed.

2. The self-reported sub-groups of Asians are evenly distributed.   

State the alternative hypothesis. Choose 1 or 2

1. The self-reported sub-groups of Asians are not evenly distributed.

2. The self-reported sub-groups of Asians are evenly distributed.    

Part (b)

What are the degrees of freedom? ________

State the distribution to use for the test Choose 1 ,2, 3 , or 4

2. t7

3. t6

Part (d)

What is the test statistic? (Round your answer to two decimal places.) __________

What is the p-value? (Round your answer to four decimal places.) ____________

Part (e)
Explain what the p-value means for this problem.

1.If H0 is true, then there is a chance equal to the p-value that the value of the test statistic will be equal to or less than the calculated value.

2.If H0is true, then there is a chance equal to the p-value that the value of the test statistic will be equal to or greater than the calculated value.  

3. If H0 is false, then there is a chance equal to the p-value that the value of the test statistic will be equal to or greater than the calculated value.

4.If H0is false, then there is a chance equal to the p-value that the value of the test statistic will be equal to or less than the calculated value.

Race Frequency Expected Frequency Asian Indian 134 ? Chinese 116 ? Filipino 1048 ? Japanese 76 ? Korean 13 ? Vietnamese 9 ? Other 23 ?

Explanation / Answer

The statistical software output for this problem is:

Chi-Square goodness-of-fit results:
Observed: Frequency
Expected: All cells in equal proportion

Hence,

a) Null hypothesis: Option 2

Alternative hypothesis: Option 1

b) Degrees of freedom = 6

c) Option 4 is correct.

d) Test statistic = 4186.28

P-value = 0.0000

e) Option 2 is correct.

N DF Chi-Square P-value 1419 6 4186.2833 <0.0001

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