95% confidence interval - proportion of youth who are not active

prop.test(x = 4404, # number of high schoolers who exercise fewer than 3 days a week
          n = 13583, # number of total trials
          p = 0.5, #null hypothesis being that the proportion of population of activity levels equal 
          alternative = "two.sided", # two-tailed alternative hypothesis
          conf.level = 0.95, # 95% confidence internal assuming normal distribution
          correct = F)

## 
##  1-sample proportions test without continuity correction
## 
## data:  4404 out of 13583
## X-squared = 1679, df = 1, p-value <2e-16
## alternative hypothesis: true p is not equal to 0.5
## 95 percent confidence interval:
##  0.316 0.332
## sample estimates:
##     p 
## 0.324

# Confidence interval = (0.316, 0.332)

Confidence Intervals - 1-sample by level of activity

#Calculate summary statistics
ci_using_formulas <- yrbss %>%
  select(weight, physical_3plus) %>%
  filter(!is.na(physical_3plus)) %>%
  group_by(physical_3plus) %>%
  summarise(mean = mean(weight, na.rm = TRUE),
            sd = sd(weight, na.rm = TRUE),
            count = n(),
            se = sd / sqrt(count),
            t = qt(0.975, count-1),
            margin = t*se,
            l_ci = mean - margin,
            h_ci = mean + margin)

 ci_using_formulas 

## # A tibble: 2 × 9
##   physical_3plus  mean    sd count    se     t margin  l_ci  h_ci
##   <chr>          <dbl> <dbl> <int> <dbl> <dbl>  <dbl> <dbl> <dbl>
## 1 no              66.7  17.6  4404 0.266  1.96  0.521  66.2  67.2
## 2 yes             68.4  16.5  8906 0.175  1.96  0.342  68.1  68.8

# Confidence interval for not active: (66.2, 67.2)
# Confidence interval for active: (68.1, 68.8)

# The confidence intervals do not overlap, and the difference should be at least 95% statistically significant.

Difference in mean hypothesis test with formula & infer

Null hypothesis: mean weights are the same for those who exercise at least 3 times a week, and those who don’t Alternative hypothesis: mean weights are different for those who exercise at least 3 times a week, and those who don’t

Test using formula:

t.test(weight ~ physical_3plus,
       alternative = "two.sided",
       conf.level = 0.95,
       data = yrbss)

## 
##  Welch Two Sample t-test
## 
## data:  weight by physical_3plus
## t = -5, df = 7479, p-value = 9e-08
## alternative hypothesis: true difference in means between group no and group yes is not equal to 0
## 95 percent confidence interval:
##  -2.42 -1.12
## sample estimates:
##  mean in group no mean in group yes 
##              66.7              68.4

# p-value very small, hence very statistically significant. We reject the null hypothesis and that there is evidence to show that there is difference in mean between those who exercise at least 3 times a week, and those who don't.

Testing using infer:

Initialise the test by calculating the means in this sample.

obs_diff <- yrbss %>%
  filter(!is.na(physical_3plus)) %>% # filter so that there are no NAs
  specify(weight ~ physical_3plus) %>%
  calculate(stat = "diff in means", order = c("yes", "no"))

obs_diff #1.77 difference in means in this sample

## Response: weight (numeric)
## Explanatory: physical_3plus (factor)
## # A tibble: 1 × 1
##    stat
##   <dbl>
## 1  1.77

Simulate the null hypothesis using infer

set.seed(1234)

null_dist <- yrbss %>%
  filter(!is.na(physical_3plus)) %>%
  # specify variables
  specify(weight ~ physical_3plus) %>%
  
  # assume independence, i.e, there is no difference
  hypothesize(null = "independence") %>%
  
  # generate 1000 reps, of type "permute"
  generate(reps = 1000, type = "permute") %>%
  
  # calculate statistic of difference, namely "diff in means"
  calculate(stat = "diff in means", order = c("yes", "no"))

Visualise and plot hypothesis testing:

ggplot(data = null_dist, aes(x = stat)) +
  geom_histogram()  

null_dist %>% visualize() +
  shade_p_value(obs_stat = obs_diff, direction = "two-sided")

null_dist %>%
  get_p_value(obs_stat = obs_diff, direction = "two_sided")

## # A tibble: 1 × 1
##   p_value
##     <dbl>
## 1       0