The variables: Type of exam and number of females
Test: Two Proportions Z-Test
\(p_1\) = Proportion of females taking the Biology Exam
\(p_2\) = Proportion of females taking the Calculus AB Exam
\(H_0\): \(p_1\) = \(p_2\)
\(H_a\): \(p_1\) > \(p_2\)
Significance Level: α = 0.05
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
The p-value is < 2.2e-16, which is very significant as it is less than α = 0.05, therefore we can reject the null.
The variables: Method Type and Crying Time
Test: Two Sample T Test
\(\mu_1\) = Average crying time of infants given shots using new methods
\(\mu_2\) = Average crying time of infants given shots using conventional methods
\(H_0\): \(\mu_1\) = \(\mu_2\)
\(H_a\): \(\mu_1\) < \(\mu_2\)
Significance Level: α = 0.05
conventional_method <- 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)
new_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(new_method, conventional_method, conf.level = 0.95, alternative = "less")
##
## Welch Two Sample t-test
##
## data: new_method and conventional_method
## 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
The p-value is 0.4881 which is greater than α = 0.05, so we fail to reject the null.