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 Numberof Households inthe CommunityPercent of U.S.Type of HouseholdHouseholdsMarried with children26%110Married, no childrenSingle parentOne personOther (e.g., roommates, siblings)29%1013225%9811%70Use 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 nul and alternate hypotheses.O Ho: The distributions are different.H: The distributions are the same.O Hg: The distributions are the same.H: The distributions are different.O Hg: The distributions are the same.H: The distributions are the same.O Hg: The distributions are different.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 YesO NoWhat sampling distribution will you use?O binomialO uniformO Student's EO chi-squareO normalWhat are the degrees of freedom?(C) Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.)(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories?O Since the P-value > a, we fall to reject the null hypothesis.O Since the P-value > a, we reject the null hypothesis.O Since the P-value sa, we reject the null hypothesis.O Since the P-value s a, we fall to reject the null hypothesis.(e) Interpret your condusion in the context of the application.O At the 5% level of significance, the evidence is sufficient to condude that the community household distribution does not fit the general U.S. household distribution.O At the 5% level of significance, the evidence is insufficient to conclude that the community household distribution does not fit the general U.S. household distribution.

Chi square χ2 test: Chi square test is useful to determine whether the proportion in each category…

The type of househald for the U.S. population and for a random sample of 411 households from a community in Montana are shown below. Observed Number Percent of U.S. of Households in Type of Household Married with children Households the Community 26% 110 Married, no children Single parent One person Other (eg., roammates, sibilings) 29% 101 32 25% 98 11% 70 Use a 5% level of significance to test the claim that the distribution of U.S. househalds fits the Dove Creek distribution. (a) What is the level of significance? State the nul and alternate hypotheses. O Ha: The distributions are different. H: The distributions are the same. O Hg: The distributions are the same. H: The distributions are different. O Hg: The distributions are the same. H: The distributions are the same. O Hg: The distributions are different. H: The distributions are different. (b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to twa decimal places. Round the test statistic to three decimal places.) Are all the expected frequencies greater than 57 O Yes O No What sampling distribution will you use? O binomial O uniform O Student's E O chi-square O normal What are the degrees of freedom? (C) Find or estimate the Pvalue of the sample test statistic. (Round your answer to three decimal places.) (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution af categories? O Since the P-value > a, we fail to reject the null hypothesis. O Since the P-value > a, we reject the null hypothesis. O Since the P-value sa, we reject the null hypothesis. O Since the P-value sa, we fall to reject the null hypothesis. (e) Interpret your condusion in the context of the application. O At the 5% level of significance, the evidence is sufficient to condude that the community household distribution does nat fit the general U.S. household distribution. O At the 5% level of significance, the evidence is insufficient to conclude that the community household distribution does not fit the general U.S. household distribution.