Hypotheses
\(H_0\): p1 <= p2
\(H_a\): p1 > p2
\(p_1\) = Is the proportion of female students that took the Biology Exam.
\(p_2\) = Is the proportion of female students that took Calculus AB Exam.
Significance Level
α = 0.05
p-value
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-value = 2.2e-16
Decision Statement:
Since the p-value < α = 0.05
There is strong evidence at the 5% significance level that the proportion of female students taking the Biology exam is higher than the proportion of female taking the Calculus AB Exam.
\(\mu1\) = The mean of babies crying time while using conventional methods.
\(\mu2\) = The mean of babies crying time while being with their mothers.
\(H_0\): \(\mu2\) = \(\mu1\)
\(H_a\): \(\mu2\) < \(\mu1\)
Significance level
α = 0.05
conventional_methods <- 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)
with_mother_method <- 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(with_mother_method, conventional_methods, alternative = "less", paired = FALSE)
##
## Welch Two Sample t-test
##
## data: with_mother_method and conventional_methods
## 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
p-value = 0.4881
Decision Statement:
We fail to reject the null hypothesis as there is not enough evidence at 5% significance to conclude that infants cry less when they are being held by their mothers in comparison to when they receive shots using conventional methods.