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.Type of HouseholdMarried with childrenMarried, no childrenSingle parentOne personOther (e.g., roommates, siblings)USE SALTPercent of U.S.Households26%Are all the expected frequencies greater than 5?O YesO NoWhat sampling distribution will you use?O chi-squareO Student's tO uniformO binomialnormal29%What are the degrees of freedom?9%25%11%Observed Numberof Households inthe Community9613129Use 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?90State the null and alternate hypotheses.O Ho: The distributions are the same. H₁: The distributions are different.65O Ho: The distributions are the same. H₁: The distributions are the same.O Ho: The distributions are different. H₁: The distributions are the same.O HO! The distributions are different. H₁: 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.) (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 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-values a, we fail to reject the null hypothesis.(e) Interpret your conclusion in the context of the application.O At the 5% level of significance, the evidence is sufficient to conclude 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.

“Since you have posted a question with multiple sub parts, we will provide the solution only to the first three sub parts as per our Q&A guidelines. Please repost the remaining sub parts separately.”Part (a) : According to the question, we have to test the hypothesis at 5% level of significance. Therefore the level of significance is given by, Level of significance = 5% or 0.05 The null and alternative hypotheses : Null hypothesis (H0) : The null hypothesis is that the distribution of U.S households fits the Dove Creek distribution. Alternative hypothesis (H1) :The alternative hypothesis is that the distribution of U.S households is different from the Dove Creek distribution. Symbolically, H0 :The distributions are the same. H1 : The distributions are different Hence, the correct option is A. Part (b) : The expected frequencies are calculated as follows : Ei = Px × ΣOi Where, ΣOi= Sum of observed frequencies. (ΣOi = 411, provided below in the table) E1 = 26% × 411 = 106.86 E2 = 29% × 411 = 119.19 E3 = 9% × 411 = 36.99 E4 = 25% × 411 = 102.75 E5 = 11% × 411 = 45.21 Using the data provided in the question we have, Percent of U.S. households (Px) Observed frequency (Oi) Expected frequency (Ei) (Oi–Ei) (Oi–Ei)2 (Oi–Ei)2/Ei 26% 96 106.86 –10.86 117.9396 1.10368 29% 131 119.19 11.81 139.4761 1.1702 9% 29 36.99 –7.99 63.8401 1.72587 25% 90 102.75 –12.75 162.5625 1.58212 11% 65 45.21 19.79 391.6441 8.66278 ΣOi = 411 Σ[(Oi–Ei)2/Ei]= 14.24465 The chi-square statistic is calculated by using the formula : χ2 = Σ[(Oi–Ei)2/Ei] Substituting the values from the table provided, χ2 = 14.24465 Rounding to the three decimal places we get, χ2 = 14.245 Chi-square statistic = 14.245 As it can be easily seen that all of the expected frequency are greater than 5. Therefore, yes all the expected frequencies are greater than 5.The chi-square distribution is used when we want to test whether the categorical variable follows the certain distribution or not. Hence, the sampling distribution is chi-square The degree of freedom is calculated as : Degree of freedom = n – 1 (Where, n = number of groups ) Degree of freedom = 5 – 1 Degree of freedom = 4 ANSWER : Chi square statistic = 14.245 Yes all the expected frequencies are greater than 5. Use chi-square distribution. Degree of freedom = 4Part (c) : Since, it is not possible to get the exact P-value using chi-square distribution table. Hence, using Excel function to calculate the P-value. P-value for chi-square statistic (i.e., 14.24) and 4 degree of freedom is as follows, Excel function : CHISQ.DIST.RT(x, df) x = test statistic, df = degree of freedom. Input : CHISQ.DIST.RT(14.245, 4) Output : P-value = 0.006552805428 Rounding to the three decimal places we get, P-value = 0.007