#    topic 17 
#  This will be remarkably similar to topics 15 and 16, so
#  much so that it would be worth comparing the three scripts.

#  First, set up the situation.  We have a population
#  with an unknown proportion of some characteristic.
#
source("../gnrnd5.R")
source("../gnrnd4.R")
gnrnd5( 38579399907, 567685) # population size is 4000 items
big_pop <- L1
#  I do not know the proportion of 1's in big_pop

#  Now, someone says that they believe that the 
#  proportion of 1's in big_pop is 20%.  
#  Thus our null hypothesis is that the proportion
#  of 1's in big_pop is 0.20. We want to look at an
#  alternative that says the proportion of 1's is 
#  greater than 0.20.  This is a one-tailed test.

#  We will get a sample of size 80 and we will
#  look at the proportion of 1's in the sample.
#  We are willing to be wrong in telling the
#  person that they are wrong 5% of the time!
#  That is, even if the true proportion of 1's in
#  big_pop is 20%, we are willing to reject the null 
#  hypothesis for 5% of the random samples that we take.
#  

########################
##  The critical value approach. We know that
##  we can use a normal approximation to the distribution
##  of sample proportions because 0.20*80 > 10.  Also,
##  our sample size is way less that 5% of the population.

##  Assuming that the null hypothesis is true, the 
##  distribution of sample proportions, p_hat, will 
##  be N( 0.20, sqrt(0.20*(1-0.20)/80))


##  This is a one-tailed test because only
##  a sample proportion that is  too high
##  would indicate that 0.20 is not the proportion.  
##  Therefore, find the value, in a normal distribution
##  that has 0.05 as the P(X > x).

# get the standard deviation
p_sd <- sqrt(0.20*(1-0.20)/80)
p_sd

#  long way

high_z <- qnorm( 0.05, lower.tail=FALSE)
high_z 
high_val <- 0.20 + high_z*p_sd
high_val


####### pause and look at a shorter way to get 
####### those two values

high_val <-qnorm( 0.05, mean=0.20, sd=p_sd,
                  lower.tail=FALSE)
high_val
##########
##########

#  Get our sample

#  the first time we do this let us get the
#  same sample each time
gnrnd4(962137901, 400000001)
L1
# take those as the index values of our random sample
our_samp <- big_pop[ L1 ]
our_samp
#  find the number of 1's in our sample
table( our_samp )
#  what proportion of the sample is that
21/80

## so now compare the proportion in the sample 
##  to our critical value.
high_val
##    In this case the sample proportion is not greater 
##    than our critical high.  Therefore, at the 
##    0.05 level of significance, we do not have evidence 
##    to reject null hypothesis that the true
##    proportion is 0.20
##    in favor of the alternative hypothesis that
##    the true proportion is greater than 0.20.

#############
#############  the attained significance approach
##   how strange would it be to get a proportion of  
##  0.2625 for a sample of size 80 if the true
##  proportion is 0.200?  
#
pnorm( 0.2625, mean=0.20, sd=p_sd, lower.tail=FALSE)
#
#  That probability is not less than our 5% level
#  of significance.  Therefore,  do not reject the 
#  null hypothesis in favor of the alternative.

#######################   Now use the function to
#######################   do the same thing
source("../hypo_prop.R")
hypoth_test_prop( 0.20, 21, 80, 1, 0.05)

#######################################
#######################################
#   Now we want to repeat this process
#   but each time we want a different sample
#   of size 80
L1 <- sample( big_pop, 80 )
freqs <- table( L1 )
freqs
num_x <- freqs[1]
hypoth_test_prop( 0.20, num_x, 80, 1, 0.05)
#### perform lines 115-119 again and again

###  now, since we have the population let us peek 
#    at the true proportion
freqs <- table( big_pop )
freqs
true_prop <- freqs[1]/4000
true_prop
#######   Try our samples again, but this time test
##        the null hypothesis that the true proportion
##        is 0.24875, and do the test at the 0.05 
##        level of significance.  
L1 <- sample( big_pop, 80 )
freqs <- table( L1 )
num_x <- freqs[1]
hypoth_test_prop( 0.24875, num_x, 80, 1, 0.05)
#### perform lines 132-135 again and again,
#### and we should see a Type I error about 
#### 5% of the time.

###  we can actually do this 1000 times and see how 
### times we reject  the null hypothesis even 
###  though it is true.
L2 <- 1:1000
for( i in 1:1000) {
  L1 <- sample( big_pop, 80 )
  freqs <- table( L1 )
  num_x <- freqs[1]
  answer <- hypoth_test_prop( 0.24875, num_x, 80, 1, 0.05) 
  L2[i] <- answer[13]
}
table( L2 )