\(H_0\): \(p_1\) = \(p_2\) \(H_a\): \(p_1\) > \(p_2\)
Where,
\(p_1\)= proportion of female students taking the biology exam \(p_2\) = Proportion of female students taking the calculus AB exam
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
Decision: Rejecting \(H_0\) at α = 0.05.
mothers <- 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)
conventional <- 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)
\(H_0\): \(\mu_1\) = \(\mu_2\) \(H_a\): \(\mu_1\) < \(\mu_2\)
Where,
\(\mu_1\) = average of infants crying when they are held by their mothers
\(\mu_2\) = average of infants crying when they held using conventional methods
t.test(mothers,conventional,alternative= "less")
##
## Welch Two Sample t-test
##
## data: mothers and conventional
## t = -0.029953, df = 57.707, p-value = 0.4881
## alternative hypothesis: true difference in means is less than 0
## 95 percent confidence interval:
## -Inf 9.135003
## sample estimates:
## mean of x mean of y
## 35.13333 35.30000
We do not reject the null. However, We do not have enough evidence to prove that average of infants crying when they are held by their mothers is less than the average of infants crying when they held using conventional methods.