Do you believe thatt an exceptionally high percentage of the executives of large corporations are right-handed? Although 85% of the general public is right-handed, a survey of 300 chief executive officers of large corporations found that 96% were right-handed.
Is this difference in percentages statistically significant? Test using \(\alpha = 0.01\).
For \(p\) the proportion of right-handed executives, we are testing \(H_0 : p = 0.85\) versus \(H_A : p > 0.85\). We have observed \(\hat{p}=0.96\) on \(n=300\). We note that \(x =\) 288 .
(p_0 <- 0.85)
## [1] 0.85
(phat <- 0.96)
## [1] 0.96
(n <- 300)
## [1] 300
(x <- n * phat)
## [1] 288
By the CLT, for \(n\) large, \(\bar{X} \sim \mbox{N}\left(p, \sqrt{p(1-p)/n}\right)\). Since \(n p_0(1-p_0) =\) 38.25 \(>5\), we can proceed.
The test statistic is \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \sim \mbox{N(0,1)} \]
For our problem we have
(z_test <- (phat - p_0)/sqrt(p_0*(1-p_0)/n))
## [1] 5.335784
To find the critical region, we need \(z_\alpha\) such that \(P(Z_\alpha > z_{1-\alpha} | H_0 \mbox{ true})\). We then check to see if the observed \(z\) is in the region \({\scr C} = \left\{z : z > z_{1-\alpha}\right\} = \mbox{RR}\).
(z_0.99 <- qnorm(0.01, lower.tail = FALSE))
## [1] 2.326348
z <- seq(-4, 6, by=0.01)
plot(z, dnorm(z), type = "l", ylab = "f(z)")
abline(v = z_0.99, col="red", lty=2)
abline(v = z_test, lty = 3)
Since \(z =\) 5.3357838 \(\in {\scr C} = \left\{z : z >\right.\) 2.3263479 \(\left. \right\}\), we reject \(H_0 : p = 0.85\) and conclude that there is sufficient evidence to suggest that the proportion of right-handed executives is greater than the proportion of right-handed people in the general population.
Find the p-value for the test and explain what it means.
The p-value is the probability that we would see a \(\hat{p}\) as, or more, extreme than we observed if \(H_0\) is actually true. In our case, the p-value is \(P(Z \ge\) 5.3357838 \()\).
(p_val <- pnorm(z_test, lower.tail = FALSE))
## [1] 4.756636e-08
Since 0 \(< \alpha = 0.01\) we reject \(H_0\) as above.
The p-value is an indicator of the probability that we would have selected one of the (in this case few) samples that would have produced a \(\hat{p}\) as large as we observed if in fact \(p = 0.85\). Since the probabilty is small, it is unlikely that we were unlucky and our assumption of \(H_0\) being true must be incorrect.