Many high school students take the AP tests in different subject areas. In 2017, of the 144,790 students who took the biology exam 84,200 of them were female. In that same year, of the 211,693 students who took the calculus AB exam 102,598 of them were female. Is there enough evidence to show that the proportion of female students taking the biology exam is higher than the proportion of female students taking the calculus AB exam? Test at the 5% level.
\(p_1\) = percentage of female
students taking AP biology
\(p_2\) = percentage of female students
taking AP calculus AB
\(H_0\): \(p_1\) = \(p_2\)
\(H_a\): \(p_1\) > \(p_2\)
\(a\) = 0.05
# p-value calculation
prop.test(c(84200, 102598), c(144790, 211693), alternative="greater")##
## 2-sample test for equality of proportions with continuity correction
##
## data: c(84200, 102598) out of c(144790, 211693)
## X-squared = 3234.9, df = 1, p-value < 2.2e-16
## alternative hypothesis: greater
## 95 percent confidence interval:
## 0.09408942 1.00000000
## sample estimates:
## prop 1 prop 2
## 0.5815319 0.4846547\(p\) = 2.2e-16
\(p\) < \(a\) thus the study is statistically
significant and the null hypothesis is rejected.
A vitamin K shot is given to infants soon after birth. The study is to see if how they handle the infants could reduce the pain the infants feel. One of the measurements taken was how long, in seconds, the infant cried after being given the shot. A random sample was taken from the group that was given the shot using conventional methods, and a random sample was taken from the group that was given the shot where the mother held the infant prior to and during the shot. Is there enough evidence to show that infants cried less on average when they are held by their mothers than if held using conventional methods? Test at the 5% level.
\(\mu_1\) = average crying with
conventional method
\(\mu_2\) = average crying with new
method
\(H_0\): \(\mu_1\) = \(\mu_2\)
\(H_a\): \(\mu_1\) < \(\mu_2\)
\(a\) = 0.05
n = c(63, 0, 2, 46, 33, 33, 29, 23, 11, 12, 48, 15, 33, 14, 51, 37, 24, 70, 63, 0, 73, 39, 54, 52, 39, 34, 30, 55, 58, 18)
a = c(0, 32, 20, 23, 14, 19, 60, 59, 64, 64, 72, 50, 44, 14, 10, 58, 19, 41, 17, 5, 36, 73, 19, 46, 9, 43, 73, 27, 25, 18)
t.test(n, a, conf.level=0.95, alternative="less")##
## Welch Two Sample t-test
##
## data: n and a
## t = 0.029953, df = 57.707, p-value = 0.5119
## alternative hypothesis: true difference in means is less than 0
## 95 percent confidence interval:
## -Inf 9.468337
## sample estimates:
## mean of x mean of y
## 35.30000 35.13333\(p\) = 0.5119
\(p\) > \(a\) thus we can not reject the null and the
test is significantly insignificant.