Statistics-WebAssign Week 8 Assignment (Homework)

QUESTION

11/16/2015Week 8 AssignmentWebAssignWeek 8 Assignment (Homework)Current Score : – / 51Due : Wednesday, November 18 2015 11:59 PM ESTYINWEI LIUCED 6030, section 02, Fall 2015Instructor: He Wang0/2 submissions1. –/1 pointsBBUnderStat11 10.1.001.In general, are chi­square distributions symmetric or skewed? If skewed, are they skewed right orleft?skewed leftsymmetricskewed rightskewed right or left2. –/1 pointsBBUnderStat11 10.1.002.For chi­square distributions, as the number of degrees of freedom increases, does any skewnessincrease or decrease? Do chi­square distributions become more symmetric (and normal) as thenumber of degrees of freedom becomes larger and larger?increases; nodecreases; yesincreases; yesdecreases; nohttps://www.webassign.net/web/Student/Assignment-Responses/last?dep=126786471/1511/16/2015Week 8 Assignment3. –/1 pointsBBUnderStat11 10.1.003.For chi­square tests of independence and of homogeneity, do we use a right­tailed, left­tailed, ortwo­tailed test?two­tailedright­tailedleft­tailedright­tailed or left­tailed4. –/1 pointsBBUnderStat11 10.1.004.In general, how do the hypotheses for chi­square tests of independence differ from those for chi­square tests of homogeneity? Explain your answer.The null for independence claims all the variables are independent whereas the null forhomogeneity claims a different proportion of interest from each population.The null for independence claims all the variables are independent whereas the null forhomogeneity claims an equal proportion of interest from each population.The null for homogeneity claims all the variables are independent whereas the null forindependence claims an equal proportion of interest from each population.The null for independence claims all the variables are dependent whereas the null forhomogeneity claims an equal proportion of interest from each population.https://www.webassign.net/web/Student/Assignment-Responses/last?dep=126786472/1511/16/2015Week 8 Assignment5. –/9 pointsBBUnderStat11 10.1.011.The following table shows site type and type of pottery for a random sample of 628 sherds at anarchaeological location.Pottery TypeMesa VerdeMcElmoMancosBlack­on­WhiteBlack­on­WhiteBlack­on­WhiteMesa Top785853189Cliff­Talus767364213Canyon Bench917362226Column Total245204179628Site TypeRow TotalUse a chi­square test to determine if site type and pottery type are independent at the 0.01 levelof significance.(a) What is the level of significance?State the null and alternate hypotheses.H0: Site type and pottery are not independent.H1: Site type and pottery are independent.H0: Site type and pottery are independent.H1: Site type and pottery are independent.H0: Site type and pottery are not independent.H1: Site type and pottery are not independent.H0: Site type and pottery are independent.H1: Site type and pottery are not independent.(b) Find the value of the chi­square statistic for the sample. (Round the expectedfrequencies to at least three decimal places. Round the test statistic to three decimalplaces.)Are all the expected frequencies greater than 5?YesNoWhat sampling distribution will you use?chi­squareStudent’s thttps://www.webassign.net/web/Student/Assignment-Responses/last?dep=126786473/1511/16/2015Week 8 AssignmentbinomialnormaluniformWhat are the degrees of freedom?(c) Find or estimate the P­value of the sample test statistic. (Round your answer to threedecimal places.)(d) Basedon your answers in parts (a) to (c), will you reject or fail to reject the nullhypothesis of independence?Since the P­value > α, we fail to reject the null hypothesis.Since the P­value > α, we reject the null hypothesis.Since the P­value ≤ α, we reject the null hypothesis.Since the P­value ≤ α, we fail to reject the null hypothesis.(e) Interpret your conclusion in the context of the application.At the 1% level of significance, there is sufficient evidence to conclude that site andpottery type are not independent.At the 1% level of significance, there is insufficient evidence to conclude that siteand pottery type are not independent.6. –/1 pointsBBUnderStat11 10.2.001.For a chi­square goodness­of­fit test, how are the degrees of freedom computed?The number of categories minus two.The number of categories plus one.The number of categories minus one.The number of categories minus three.https://www.webassign.net/web/Student/Assignment-Responses/last?dep=126786474/1511/16/2015Week 8 Assignment7. –/1 pointsBBUnderStat11 10.2.002.How are expected frequencies computed for goodness­of­fit tests?Divide the total sample size by the sample size for each category.Take the proportion of the sample size for each category designated by the proposeddistribution.Divide the proportion of the sample size for each category by the total sample size.Take the proportion of the sample size for each category from the observed data.8. –/1 pointsBBUnderStat11 10.2.003.Explain why goodness­of­fit tests are always right­tailed tests.We use a χ2 distribution where only smaller values can lead to rejecting the null.We use a binomial distribution where only larger values can lead to rejecting the null.We use a χ2 distribution where only larger values can lead to rejecting the null.We use a normal distribution where only larger values can lead to rejecting the null.https://www.webassign.net/web/Student/Assignment-Responses/last?dep=126786475/1511/16/2015Week 8 Assignment9. –/1 pointsBBUnderStat11 10.2.004.When the sample evidence is sufficient to justify rejecting the null hypothesis in a goodness­of­fittest, can you tell exactly how the distribution of observed values over the specified categoriesdiffers from the expected distribution? Explain your answer.Yes. When we reject the null, we can only conclude that the observed distribution isdifferent from the expected distribution.No. When we reject the null, we can only conclude that the observed distribution is differentfrom the expected distribution.Yes. When we reject the null, we can tell exactly how the observed distribution is differentfrom the expected distribution.No. When we reject the null, we can tell exactly how the observed distribution is differentfrom the expected distribution.https://www.webassign.net/web/Student/Assignment-Responses/last?dep=126786476/1511/16/2015Week 8 Assignment10.–/9 pointsBBUnderStat11 10.2.006.The type of household for the U.S. population and for a random sample of 411 households from acommunity in Montana are shown below.Observed NumberPercent of U.S.Type of Householdof Households inHouseholdsthe CommunityMarried with children26%104Married, no children29%1189%30One person25%90Other (e.g., roommates, siblings)11%69Single parentUse a 5% level of significance to test the claim that the distribution of U.S. households fits theDove Creek distribution.(a) What is the level of significance?State the null and alternate hypotheses.H0: The distributions are the same.H1: The distributions are the same.H0: The distributions are the same.H1: The distributions are different.H0: The distributions are different.H1: The distributions are the same.H0: The distributions are different.H1: The distributions are different.(b) Find the value of the chi­square statistic for the sample. (Round the expectedfrequencies to two decimal places. Round the test statistic to three decimal places.)Are all the expected frequencies greater than 5?YesNoWhat sampling distribution will you use?Student’s thttps://www.webassign.net/web/Student/Assignment-Responses/last?dep=126786477/1511/16/2015Week 8 Assignmentbinomialuniformnormalchi­squareWhat are the degrees of freedom?(c) Find or estimate the P­value of the sample test statistic. (Round your answer to threedecimal places.)(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the nullhypothesis that the population fits the specified distribution of categories?Since the P­value > α, we fail to reject the null hypothesis.Since the P­value > α, we reject the null hypothesis.Since the P­value ≤ α, we reject the null hypothesis.Since the P­value ≤ α, we fail to reject the null hypothesis.(e) Interpret your conclusion in the context of the application.At the 5% level of significance, the evidence is sufficient to conclude that thecommunity household distribution does not fit the general U.S. householddistribution.At the 5% level of significance, the evidence is insufficient to conclude that thecommunity household distribution does not fit the general U.S. householddistribution.https://www.webassign.net/web/Student/Assignment-Responses/last?dep=126786478/1511/16/2015Week 8 Assignment11.–/1 pointsBBUnderStat11 10.3.001.Does the x distribution need to be normal in order to use the chi­square distribution to test thevariance? Is it acceptable to use the chi­square distribution to test the variance if the xdistribution is simply mound­shaped and more or less symmetric?Yes, it needs to be normal. No, the chi­square test of variance requires the x distribution tobe exactly normal.No, it does not need to be normal. Yes, the chi­square test of variance allows for the xdistribution to be simply mound­shaped or symmetric.Yes, it needs to be normal. Yes, the chi­square test of variance allows for the x distributionto be simply mound­shaped or symmetric.No, it does not need to be normal. No, the chi­square test of variance requires the xdistribution to be exactly normal.https://www.webassign.net/web/Student/Assignment-Responses/last?dep=126786479/1511/16/2015Week 8 Assignment12.–/11 pointsBBUnderStat11 10.3.004.Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941,the population variance of x was approximately σ2 = 5.1. Suppose a recent study of age at firstmarriage for a random sample of 31 women in rural Quebec gave a sample variance s2 = 2.5. Usea 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90%confidence interval for the population variance.(a) What is the level of significance?State the null and alternate hypotheses.Ho: σ2 = 5.1; H1: σ2 ≠ 5.1Ho: σ2 < 5.1; H1: σ2 = 5.1Ho: σ2 = 5.1; H1: σ2 > 5.1Ho: σ2 = 5.1; H1: σ2 < 5.1(b) Find the value of the chi­square statistic for the sample. (Use 2 decimal places.)What are the degrees of freedom?What assumptions are you making about the original distribution?We assume a binomial population distribution.We assume a uniform population distribution.We assume a exponential population distribution.We assume a normal population distribution.(c) Find or estimate the P­value of the sample test statistic. (Use 4 decimal places.)(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the nullhypothesis?At the α = 0.05 level, we reject the null hypothesis and conclude the data are notstatistically significant.At the α = 0.05 level, we reject the null hypothesis and conclude the data arestatistically significant.https://www.webassign.net/web/Student/Assignment-Responses/last?dep=1267864710/1511/16/2015Week 8 AssignmentAt the α = 0.05 level, we fail to reject the null hypothesis and conclude the data arenot statistically significant.At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data arestatistically significant.(e) Interpret your conclusion in the context of the application.Fail to reject the null hypothesis, there is sufficient evidence to conclude thevariance of age at first marriage is less than 5.1.Reject the null hypothesis, there is insufficient evidence to conclude the variance ofage at first marriage is less than 5.1.Reject the null hypothesis, there is sufficient evidence to conclude the variance ofage at first marriage is less than 5.1.Fail to reject the null hypothesis, there is insufficient evidence to conclude thevariance of age at first marriage is less than 5.1.(f) Find the requested confidence interval for the population variance or populationstandard deviation. (Use 2 decimal places.)lower limitupper limitInterpret the results in the context of the application.We are 90% confident that σ2 lies outside this interval.We are 90% confident that σ2 lies above this interval.We are 90% confident that σ2 lies below this interval.We are 90% confident that σ2 lies within this interval.https://www.webassign.net/web/Student/Assignment-Responses/last?dep=1267864711/1511/16/2015Week 8 Assignment13.–/1 pointsBBUnderStat11 10.4.001.When using the F distribution to test variances from two populations, should the random variablesfrom each population be independent or dependent?independentdependent14.–/1 pointsBBUnderStat11 10.4.002.When using the F distribution to test two variances, is it essential that each of the two populationsbe normally distributed? Would it be all right if the populations had distributions that were mound­shaped and more or less symmetrical?Yes, both populations must be normal. Mound­shaped or symmetric distributions willqualify.No, the populations do not have to be normal. Mound­shaped or symmetric distributions donot qualify.No, the populations do not have to be normal. Mound­shaped or symmetric distributions willqualify.Yes, both populations must be normal. Mound­shaped or symmetric distributions do notqualify.15.–/1 pointsBBUnderStat11 10.4.003.In general, is the F distribution symmetrical? Can values of the F distribution be negative?Yes; NoYes; YesNo; NoNo; Yeshttps://www.webassign.net/web/Student/Assignment-Responses/last?dep=1267864712/1511/16/2015Week 8 Assignment16.–/1 pointsBBUnderStat11 10.4.004.To use the F distribution, what degrees of freedom need to be calculated?Only the degrees of freedom for the numerator need to be calculated.Only the degrees of freedom for the denominator need to be calculated.The degrees of freedom for both the numerator and denominator need to be calculated.https://www.webassign.net/web/Student/Assignment-Responses/last?dep=1267864713/1511/16/2015Week 8 Assignment17.–/9 pointsBBUnderStat11 10.4.012.A new thermostat has been engineered for the frozen food cases in large supermarkets. Both theold and new thermostats hold temperatures at an average of 25°F. However, it is hoped that thenew thermostat might be more dependable in the sense that it will hold temperatures closer to25°F. One frozen food case was equipped with the new thermostat, and a random sample of 26temperature readings gave a sample variance of 5.1. Another similar frozen food case wasequipped with the old thermostat, and a random sample of 19 temperature readings gave asample variance of 12.2. Test the claim that the population variance of the new thermostattemperature readings is smaller than that for the new thermostat. Use a 5% level of significance.How could your test conclusion relate to the question regarding the dependability of thetemperature readings? (Let population 1 refer to data from the old thermostat.)(a) What is the level of significance?State the null and alternate hypotheses.H0: σ12 = σ22; H1: σ12 > σ22H0: σ12 = σ22; H1: σ12 ≠ σ22H0: σ12 = σ22; H1: σ12 < σ22H0: σ12 > σ22; H1: σ12 = σ22(b) Find the value of the sample F statistic. (Round your answer to two decimal places.)What are the degrees of freedom?dfN =dfD =What assumptions are you making about the original distribution?The populations follow dependent normal distributions. We have random samplesfrom each population.The populations follow independent normal distributions. We have random samplesfrom each population.The populations follow independent chi­square distributions. We have randomsamples from each population.The populations follow independent normal distributions.https://www.webassign.net/web/Student/Assignment-Responses/last?dep=1267864714/1511/16/2015Week 8 Assignment(c) Find or estimate the P­value of the sample test statistic. (Round your answer to fourdecimal places.)(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the nullhypothesis?At the α = 0.05 level, we reject the null hypothesis and conclude the data are notstatistically significant.At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data arestatistically significant.At the α = 0.05 level, we reject the null hypothesis and conclude the data arestatistically significant.At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data arenot statistically significant.(e) Interpret your conclusion in the context of the application.Reject the null hypothesis, there is sufficient evidence that the population varianceis smaller in the new thermostat temperature readings.Fail to reject the null hypothesis, there is insufficient evidence that the populationvariance is smaller in the new thermostat temperature readings.Fail to reject the null hypothesis, there is sufficient evidence that the populationvariance is smaller in the new thermostat temperature readings.Reject the null hypothesis, there is insufficient evidence that the population varianceis smaller in the new thermostat temperature readings.https://www.webassign.net/web/Student/Assignment-Responses/last?dep=1267864715/15

ANSWER:

REQUEST HELP FROM A TUTOR