Percent of U.S. of Households in Households 26% 29% 9% 25% 11% pe of Household t

Question

Percent of U.S. of Households in Households 26% 29% 9% 25% 11% pe of Household the Community 109 arried with children Married, no children Single parent One person Other (e.g., roommates, siblin 106 72 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 Ho: The distributions are different. H: The distributions are the same O Ho: The distributions are different. H1: The distributions are different. O Ho: The distributions are the same H1: The distributions are the same O Ho: The distributions are the same H: The distributions are different. (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? O Yes O No

Explanation / Answer

(a)The level of significance is 5%.

(B)The null and alternative hypotheses are:

H0:The distributions are the same

H1:The distributions are different.

The degrees of freedom=k-1=5-1=4

Now expected value is Ei=n*pi

E1=411*0.26=106.86

E2=411*0.29=119.19

E3=411*0.09=36.99

E4=411*0.25=102.75

E5=411*0.11=45.21

The test statistic chi square is:

Sum over(Oi-Ei)^2/Ei

=(109-106.86)^2/106.86+…+(72-45.21)^2/45.21

=19.15

C.The P value is determined using a P value calculator: 0.000734.

d.The P value is less than 0.05. Thus reject the null hypothesis.

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