Hello! I'm wondering if the same tutor that did order #100534will be able to work on this...

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Hello! I'm wondering if the same tutor that did order #100534will be able to work on this one.Microsoft Word - 문서1   1. A random sample of 200 people is drawn and their political affiliations and their opinions of the  congress are recorded in the table below.     Opinions of congress row  Political Affiliation Approve Disapprove Net Total  Conservative 10 (?) 50 20 80  Liberal 20 10 10 40  Moderate 20 30 30 80  Column Total 50 90 60 200    We are interested in knowing whether their opinions are associated with their political affiliations. W hich of the following test can be conducted?  1) Chi-Square test for the goodness of fit  2) Chi-Square test of homogeneity  3) Two sample test  4) Chi-square test of independence          2. (cont.) What is the expected frequency of the cell with a question mark?   1) 80*50/200=20   2) 10   3) 50*0.25=12.5   4) None of the above          3. (cont.) What is the degree of freedom associated with the test statistic    ! {#0!" − ?!"' # !" /?!")}?  1) 2*3  2) 2+3  3) 3*3  4) 2*2        4. A 1998 study reported that the average weight of newborn infants is 7 pounds. You take a simple  random of 16 newborns to see if the average birth weight is increased. Suppose the sample mean  is 7.3 pounds and the sample standard deviation is 0.8 pound. Assume that the birth weight approxi mately follows a normal distribution.   1) ?$:	? = 7		??. ?%:	? ≠ 7   2) ?$:	? = 7.3		??. ?%:	? ≠ 7.3   3) ?$:	? = 7.3		??. ?%:	? > 7.3   4) ?$:	? = 7		??. ?%:	? > 7          5. (cont.) The appropriate test statistic is:   1) &'&.) *$.+/-.    2) &.) $.+/√-.    3) &.)'& $.+/√-.    4) &.)'& *$.+/-.           6. (cont.) Suppose the test stat is stored in an R variable named teststat. What is the R code to co mpute   1) 1-pt(teststat, df=16)   2) 1-pnorm(teststat)   3) 2*(1-pt(teststat, df=15))   4) 1-pt(teststat, df=15)          7. (cont.) Suppose the p-value is 0.777. What conclusion should we draw at the significance level ? = 0.05?   1) Accept ?% because p-value>0.05   2) Do not reject ?$ because p-value >0.05   3) Reject ?$ because p-value>0.05   4) None          8. Suppose we would like to compute the means of two populations. We consider the following test:  (1) two-sample t (2) two-sample z (3) chi-square test of goodness of fit (4) chi-square test of indep endence. If we draw two independent simple random from two populations, which of the following t est may be conducted?   1) 1,2   2) 2,3   3) 3,4   4) 1,2,3          9. Let the explanatory variable X and the response variable be Y. The least square regression equatio n is Y=1.34+0.35X. The correlation coefficient between X and Y is   1) 0   2) 0.35   3) between 0 and 1   4) None          10. Some people think that chemist factors are more likely than others to have female children. The  Washington State Department of Health list the parents’ occupation on birth certificates. Suppose bet ween 6-month period, 55 children were born to fathers who were chemists. Of these births, 27 were  girls. During the same period, 48.8% of all births in Washington were girls. Let p denote the propor tion of chemist fathers with female children.   1) ?$:	? = 0.488	??. ?-:	? ≠ 0.488   2) ?$:	? = 0.488		??. ?-: ? > 0.488   3) ?$:	? = 0.488		??. ?-:	?  str(cars)  $ Price   : num[1:804]  17.3 17.5 16.2 16.337 16.3 ...   $ Mileage : num[1:804]    8221 9135 13196 16342 19832  ...   $ Make    : Factor w/ 6 levels "Buick","Cadillac",..: 1 1 1 1 1 1 1 1 1 1 ...  >    # We begin by fitting the following model to the data:  > fit1 summary(fit1)    Call:  lm(formula = Price ~ Mileage+Make, data=cars)    Residuals:     Min     1Q       Median     3Q    Max   -11.7552  -3.2740  -0.7018   1.517 28.1741      Coefficients:                  Estimate.    Std. Error   t value   Pr(>|t|)      (Intercept)    2.431e+01   8.182e-01     29.7      fit0anora (fit0, fit1)    Analysis of variance table    Model1: Price ~1  Model2: Price ~ Mileage +Make    Res.DF     RSS       Df   Sum of Square    F       Pr>F   1  803         78461  2   797        76312     (?)    (??)           26228

Atreye · Accepted Answer

Solution 1:
4) Chi-square test of independence.
Solution 2:
1) 80*50/200 = 20
Solution 3:
4) 2*2
Solution 4:
4) 
Solution 5:
3) 
Solution 6:
4)1-pt(teststat, df=15)
Solution 7:
2) Do not reject  because p-value >0.05
Solution 8:
1) 1,2 
Solution 9:
3) between 0 and 1
Solution 10:

Microsoft Word - 문서1 1. A random sample of 200 people is drawn and their political affiliations and their opinions of the congress are recorded in the table below. Opinions of congress row Political...

Answer To: Microsoft Word - 문서1 1. A random sample of 200 people is drawn and their political affiliations and...

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