Question

The type of household for the U.S. population and for a random sample of 411 households from a community in Montana are shown below. Observed Number of Households in the Community 90 Type of Household Married with children Married, no children Single parent One person Other (e.g., roommates, siblings) Percent of U.S. Households 26% 29% 9% 25% 11% 126 28 100 67 Use a 5% level of significance to test the claim that the distribution of U.S. households fits the Dove Creek distribution. (a) What is the level of significance? State the null and alternate hypotheses. O H : The distributions are different. H,: The distributions are different. OH,: The distributions are the same. H,: The distributions are the same. O Ho: The distributions are the same. H,: The distributions are different. OH: The distributions are different. H: The distributions are the same. (b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to two decimal places. Round the test statistic to three decimal places.) Are all the expected frequencies greater than 5? Yes O No What sampling distribution will you use? uniform chi-square binomial Student's t O normal What are the degrees of freedom?

Answer 1

proportions Observed number of household (0) Expected E=411*p Chi square =(O-E)2/E (p) Type of Household Married with children Married no children Single parent one person Other 90 126 28 100 67 0.26 0.29 0.09 0.25 0.11 106.86 119.19 36.99 102.75 45.21 2.6601 0.3891 2.1849 0.0736 10.5022 411 15.8099 a) Level of significance a = 0.05 Test hypothesis are Ho:The distributions are same H:The distributions are different

b) Chi-square test statistic x2 =2(0-5)) = 15.810 all expected values are > 5 Yes With sampling distribution we use Chi-square distribution Degrees of freedom df=n-1 = 5-1 = 4 c) p-value =CHIDIST(15.810 ,4)=0.0004 =0.000 d) Since p-value < 0.05 hence reject the null hypothesis 3rd option is correct e) 1st option is correct