Question

A recent national report states the marital status distribution of the male population age 18 or older is as follows: Never Married (31.7%), Married (54.8%), Widowed (3%), Divorced (10.5%). The table below shows the results of a random sample of 1768 adult men from California. Test the claim that the distribution from  California is as expected at the αα = 0.05 significance level.

1. Complete the table by filling in the expected frequencies. Round to the nearest whole number:

   Frequencies of Marital Status

   Outcome	Frequency	Expected Frequency
   Never Married	578
   Married	974
   Widowed	34
   Divorced	182

   R vector Data: 578,974,34,182
   Excel Data
   
   Expected Frequency Never Married =
   Expected Frequency Married =  
   Expected Frequency Widowed =
   Expected Frequency Divorced =

1. What is the correct statistical test to use?
   Select an answer Paired t-test Homogeneity Goodness-of-Fit Independence  
2. What are the null and alternative hypotheses?
   H0:H0: Select an answer Marital status and residency are dependent. The distribution of marital status in California is the same as it is nationally. Marital status and residency are independent. The distribution of marital status in California is not the same as it is nationally.  
   
   
   H1:H1: Select an answer Marital status and residency are independent. The distribution of marital status in California is the same as it is nationally. The distribution of marital status in California is not the same as it is nationally. Marital status and residency are dependent.
3. The degrees of freedom =
   
   
4. The test-statistic for this data = (Please show your answer to three decimal places.)
   
   
5. The p-value for this sample =(Please show your answer to four decimal places.)
   
   
6. The p-value is Select an answer less than (or equal to) greater than  αα  
   
   
7. Based on this, we should Select an answer reject the null accept the null fail to reject the null  
   
   
8. Thus, the conclusion is... Select an answer At the 5% significance level, there is insufficient evidence to conclude that the distribution of marital status in California is not the same as it is nationally. At the 5% significance level, there is sufficient evidence to conclude that the distribution of marital status in California is not the same as it is nationally.

Answer 1

Categories Observed Expected (fo-fe)?/fe Category 1 578 1768+0.317=560.456 (578-560.456) 2/560.456 0.549 Category 2 974 1768*0.548=968.864 (974-968.864)2/968.864 0.027 Category 3 34 1768*0.03=53.04 (34-53.04)2753.04 = 6.83 Category 4 182 1768*0.105=185.64 (182-185.64)2/185.64 = 0.071 Sum = 1768 1768 7.483 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: H: P1 = 0.317, P2 = 0.548, P3 = 0.03, P4 = 0.105 He: Some of the population proportions differ from the values stated in the null hypothesis This corresponds to a Chi-Square test for Goodness of Fit. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, the number of degrees of freedom is df = 4-1 =3, so then the rejection region for this test is R= {x?:X? > 7.815} (3) Test Statistics The Chi-Squared statistic is computed as follows: x² = (0; – E;)? E; = 0.549 + 0.027 + 6.835 + 0.071 7.483 i=1 (4) Decision about the null hypothesis Since it is observed that x? = 7.483 < xả = 7.815, it is then concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the a = 0.05 significance level.

Chi-square = 7.483, df = 3 Right-tail p-value is 0.058 Rcommand: pchisq(7.483, 3, lower.tail=FALSE) pchisq(7.483,3) or 0 2 4 6 8 10 12 14

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