Problem 6.6

  1. False. We know that 46% of our poll sample supports the decision.
  2. True. The poll indicates that we with 95% confidence can claim that between 43% and 49% of Americans support the decision.
  3. True. Our sample proportion distribution should be normally distributed with 95% samples having the proportion between 43% and 49%.
  4. False.As our confidence level goes down confidence interval shrinks. We have better more narrow interval but we also less confident about the interval.

Problem 6.28

Point estimate +-z*SE, where point estimate is 0.088-0.08=0.008

z for 95% confidence interval is 1.96.

and SE =sqrt(SE12+SE22)

The answer is (0.0079964,0.0080358)

SE1<-0.088*(1-0.088)/4691

SE2<-0.08*(1-0.08)/11545

SE<-sqrt(SE1**2+SE2**2)

0.008-1.96*SE
## [1] 0.007964215
0.008+1.96*SE
## [1] 0.008035785

Problem 6.44

  1. H0) Deer proportions for all habitats are the same as proportion of habitats. H1) Deer proportions for some of habitats are not the same as proportions of habitats.

  2. Chi-square statistics

  3. Condtitions to check: Independence - the observations look independent. Expected Values for each group should be 5 or more. The condition is satisfied. For wood expected value iis 20.448

  4. Chi-square statistics equal (4-4260.048)2/(4260.048)+(16-4260.1426)2/(4260.1426)+(67-426*0.396)**2/(4260.396)+(345-426(1-0.048-0.1426-0.396))/(426*(1-0.048-0.1426-0.396)) or 108.4572. Degrees of freedom is 3 degree of freedom (4 backets - 1). Such a high chi-squire value for just 3 degree of freedom indicatees very low p-value (less than 0.001), so null hypotesis is rejected. Deer distribution is not proportional to habitat distribution. (Actually for wood we have very big discrepency 4 (actual) vs 20 (expected value))

EV1<-426*0.048

EV1
## [1] 20.448
(4-426*0.048)**2/(426*0.048)+(16-426*0.1426)**2/(426*0.1426)+(67-426*0.396)**2/(426*0.396)+(345-426*(1-0.048-0.1426-0.396))/(426*(1-0.048-0.1426-0.396))
## [1] 108.4572