title: “HW 2 DACSS Eris Dodds” format: html: theme: default

Question 1

See Code

bypass_mean<-19
bypass_sd<-10
bypass_size<-539
standard_error_bypass<-bypass_sd/sqrt(bypass_size)
confidence_level<-0.9
tail_area<-(1-confidence_level)/2
t_score<-qt(p = 1-tail_area, df = bypass_size-1)
CI_bypass<-c(bypass_mean - t_score * standard_error_bypass, bypass_mean + t_score * standard_error_bypass)
print(CI_bypass)
[1] 18.29029 19.70971
anio_mean<-18
anio_sd<-9
anio_size<-847
standard_error_anio<-anio_sd/sqrt(anio_size)
tail_area_anio<-(1-confidence_level)/2
t_score_anio<-qt(p = 1-tail_area_anio, df = anio_size)
CI_anio<-c(anio_mean - t_score_anio * standard_error_anio, anio_mean + t_score_anio * standard_error_anio)
print(CI_anio)
[1] 17.49078 18.50922

Question 2

PE<-.55 Lower Limit = .52 Upper Limit = .58

pop<-567
size<-1031
PE<-pop/size
prop.test(pop,size)

    1-sample proportions test with continuity correction

data:  pop out of size, null probability 0.5
X-squared = 10.091, df = 1, p-value = 0.00149
alternative hypothesis: true p is not equal to 0.5
95 percent confidence interval:
 0.5189682 0.5805580
sample estimates:
        p 
0.5499515

Question 3

(1.96)^2 *(42.5)^2/ (5)^2
[1] 277.5556

Question 4

The p value is .049, showing it is statisitcally significant and we can reject the null. Other aspects to this question are verified within the code

fem_size<-9
fem_mean<-410
null<-500
sd<-90

SE<-sd/sqrt(fem_size)
t_score<-(fem_mean-null)/SE
p_value<-(pt(t_score, df=9))*2
upper_p<-(pt(t_score, df=8, lower.tail = FALSE))
upper_p
[1] 0.9914642
lower_p<-(pt(t_score, df=8, lower.tail = FALSE))
lower_p
[1] 0.9914642

Question 5

  1. see code
  2. Jones = .051 not significant, cannot reject null. Smith = .049 significant, reject null.
  3. Being broad about the direction of the p value, in this case, would overshadow how marginally significant and insignificant the p values actually came out to in this case.
jones_mean<-519.5
smith_mean<-519.7
jones_se<-10
smith_se<-10
null<-500
jones_t<-(jones_mean-null)/jones_se
jones_t
[1] 1.95
smith_t<-(smith_mean-null)/smith_se
smith_t
[1] 1.97
jones_p<-pt(jones_t, df=999, lower.tail = FALSE)*2
jones_p
[1] 0.05145555
smith_p<-pt(smith_t, df=999, lower.tail = FALSE)*2
smith_p
[1] 0.04911426

Question 6

The results of a t test show that the mean is less that 45, with a relatively small p value. We can have more confidence, then, that the average gas tax per gallon was less than .45 cents. ::: {.cell}

gas_taxes <- c(51.27, 47.43, 38.89, 41.95, 28.61, 41.29, 52.19, 49.48, 35.02, 48.13, 39.28, 54.41, 41.66, 30.28, 18.49, 38.72, 33.41, 45.02)
t.test(gas_taxes, mu = 45, alternative = "less")

    One Sample t-test

data:  gas_taxes
t = -1.8857, df = 17, p-value = 0.03827
alternative hypothesis: true mean is less than 45
95 percent confidence interval:
     -Inf 44.67946
sample estimates:
mean of x 
 40.86278

:::