6.6 2010 Healthcare Law

• 1. No, we are 100% sure that 46% of people in this sample support the decision
• 2. Yes, our confidence interval is extrapolated from our sample data and we are 95% sure the average population mean is within that Confidence interval
• 3. No. 95% of our samples confidence intervals will contain the mean, however, we can’t tell how many sample means may fall outside of this samples means confidence interval
• 4. False, as we decrease the confidence level requirement, the margin of error decreases

6.12 Legalization of marijuana, Part I.

• 1. It is a sample statistic. Although it can be used to estimate our population parameter, it is the average of our sample.
• • N= 1,259

  • z=1.96

  • PE= 48

  • SE = \(\sqrt {\frac {P_0(1-P_0)}{N}}\)= \(\sqrt {\frac {.5(1-.5)}{1,259}}\)

  • CI= point estimate \(\pm\) Z * SE

  • Our confidence interval is around 45-50%. Contextually that means the public is pretty much split on support of legalization

se <-  sqrt((.5)*.5/1259)
paste("Se= ",round(se,3))

## [1] "Se=  0.014"

upper <- .48+ 1.96*se
lower <- .48-1.96*se
paste("our confidence interval is",lower,"-",upper)

## [1] "our confidence interval is 0.452380665449998 - 0.507619334550002"

• C. to use normal model two conditions must be met
  • Independence- less than 10% population
  • Success-failure conditions must be satisfied.
  • Both conditions are met, thus I would consider the use of the normal model valid in this example
• D. Considering 50% is within our confidence interval it is possible. But stating it as a fact is misleading

6.20 Legalize Marijuana, Part II

\(N= Z^2*.48*.52/ ME^2\)

• answer= 2398

n=1.96^2*.48*.52/.02^2
paste (n)

## [1] "2397.1584"

6.28 Sleep deprivation, CA vs. OR, Part I.

• 1. N_1= 11,545

     P_1=.08

     N_2= 4691

     P_2= .088

     z=1.96

  PE= \(p_1-p_2\)= .8

  SE = \(\sqrt {\frac {P_1(1-P_1)}{N_1}+\frac {P_2(1-P_2)}{N_2}}\)= \(\sqrt {\frac {.5(1-.5)}{1,259}}\)

  CI= point estimate \(\pm\) Z * SE

N_1= 11545
P_1=.08
N_2= 4691
P_2= .088
z=1.96

se <-  sqrt( ((.08)*(1-.08)/11545)+((.088)*(1-.088)/4691))
upper <- .008+ 1.96*se
lower <- .008-1.96*se
paste("our confidence interval is",round(lower*100,3),"% to",round(upper*100,3),"%")

## [1] "our confidence interval is -0.15 % to 1.75 %"

• Contextually this means that random sampling can account for the differences in means that were observed between these samples

6.44 Barking Deer

percents <- c(.048,.147,.396,(1-.048-.147-.396))
null <- round(percents*426,0)
print (null)

## [1]  20  63 169 174

• A. \(H_o\) Barking deer don’t prefer to forage in certain habitats

  \(H_A\) Barking deer prefer to forage in certain habitats

• 2. Chi square goodness of fit

C. conditions are as follows

• Sampling method is simple random sampling.- True

• The variable under study is categorical.- True

• The expected value of the number of sample observations in each level of the variable is at least 5.- True

• z= \(\frac{observed- null}{SE_o}\)

• |Z1| + |Z2| + |Z3|+|z4|

percents <- c(.048,.147,.396,(1-.048-.147-.396))
actual <- c(4,16,67,345)
null <- round(percents*426,0)
deers = as.data.frame(rbind(actual, null))
names(deers) = c('Woods','grass','Deciduous','Other') 
short_hand_x_squared <- chisq.test(deers,correct = TRUE)

long_hand_x_squared <- sum((actual-null)^2/null)

paste("with predefined function",short_hand_x_squared[1], " chi^2 by hand", long_hand_x_squared)

## [1] "with predefined function 139.019318509426  chi^2 by hand 277.477346378938"

• For some reason the chi square comes out differently here, but both indicate that there Barking deer prefer to forage in certain habitats

6.48 Coffee and Depression

• A. Chi square test for two way tables

• B. \(H_o\) there is no association between coffee intake and depression

\(H_A\) there is an association between coffee intake and depression

• C. total depression ratio= 5.13% of women suffer from depression

2607/50739

## [1] 0.05138059

• 4. expected_count= 339 Chi_square= 3.3

expected <- 6617*.0513
actual <- 373

chi_sq <- (actual-expected)^2/expected

paste("expected value count is ",expected," the chi square value is ",chi_sq)

## [1] "expected value count is  339.4521  the chi square value is  3.31552402948752"

• E. Using the chi square probability table the P value is less than .001 and using R it comes out to .0003

C <- 5
R <- 2

df <- (R-1)*(C-1)
chi2 <- 20.93

p_value <- 1 - pchisq(chi2, df)

• F. The conclusion of the study is we can reject the null hypothesis and state there is an association between coffee intake and depression

• 7. As this was an observational study, causation can’t be concluded. Therefore I agree it is too early to recommend women load up on coffee