The type of household forthc u.s. pcpulation and for ? iadom sampie of 411 hcuse
ID: 3054717 • Letter: T
Question
The type of household forthc u.s. pcpulation and for ? iadom sampie of 411 hcusenolds from ? community In Montana anc shown below. Obscrved Number Percent of U.S, Househulds i Type ot Household Harriad with childran Married, no childrer Single parent 112 25 Use a S? laval of slaincanca to test the claim that the distribution at U.S. housahclcs tits ??? Doe Creek distrioutlan. State tha null and aet hipothcsas. ) HO: The distributions ana the same. H:The distributions are dittcrant. H: The distrbutionc are ciffcrent. rfu; "'e ?.sribution, dr , ?.rrerenl. HTha dbutians are the sama. Hg: The diztributions are the 2anc H: The distributions are the same b rd the value or the, Uhi squ”? statistic r r the sar?ple. Ruund the expected rrequenuts to two de anal pla. o Round the test Latistic tu thr deurmal tl Uws Are all the expected frequencics greater than 5 what sampling distribution will you use? binomial chi squae studentst narmal unifomm what arc the degrecs of freedom Find ar estimat the value of the sample test statistic. (Round your awer to three decimal places.)Explanation / Answer
a) level of significance =0.05
Ho: the distribution are same
H1: the distribution are different
b)applying chi square goodness of fit:
test statistic =19.552
all expected frequency greater than 5:--Yes
sampling distirbution: chi square
degree of freedom =categories-1=5-1 =4
c) estimated p value =0.001
observed Expected Chi square category Probability(p) Oi Ei=total*p R2i=(Oi-Ei)2/Ei married; with children 0.260 101.000 106.86 0.32 married; no children 0.290 112.000 119.19 0.43 single parent 0.090 31.000 36.99 0.97 one person 0.250 94.000 102.75 0.75 other 0.110 73.000 45.21 17.08 total 1 411 411 19.5524Related Questions
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