Homework #6

Problem 10.12

Since y = 0, we estimate the adjusted \(\pi\)

• \(\hat{\pi}_{Adj} = \frac{3/8}{n+3/4} = \frac{3/8}{20+3/4}\) = 0.01807229
• 95% CI = \([0, 1-(\alpha/2)^{1/n}]\) = \([0, 1-(0.05/2)^{1/20}]\) = [0, 0.1684335]

Problem 10.14

• (a). Yes. Because both \(n\pi \geq 5\) and \(n(1-\pi) > 5\)

• (b). Yes. p-value = 0.04829 < 0.05. The null hypothesis is rejected.

  • \(H_{0}: p \leq 0.5\)

  • \(H_{a}: p > 0.5\)

  • calculation:
    • \(\hat{\pi}\) = 424/800 = 0.53
    • \(\pi_{0}\) = 0.5
    • \(\sigma_{\hat{\pi}} = \sqrt{\frac{\pi_{0}(1-\pi_{0})}{n}}\) = \(\sqrt{0.5(1-0.5)/800}\) = 0.01767767
    • \(z = \frac{\hat{\pi} - \pi_{0}}{\sigma_{\hat{\pi}}}\) = 1.697056
    • \(z = 1.697056 > z_{0.95} = 1.644854\)

• (c). 95% CI: \(\hat{\pi} \pm 1.96\sigma_{\hat{\pi}}\) = \(0.53 \pm 0.01767767\) = [0.5123223, 0.5476777]

Problem 10.18

• parameter estimates

  • \(\mu = \hat{\pi_{1}} - \hat{\pi_{2}} = 91/250 - 53/250 = 38/250 = 0.152\)
  • \(\sigma = \sqrt{\frac{\pi_{1}(1-\pi_{1})}{n1} + \frac{\pi_{2}(1-\pi_{2})}{n2}}\) = 0.03992794
  • \(z = \mu/\sigma\) = 0.152/0.03992794 = 3.806858

p1 = 91/250
p2 = 53/250
n1 = 250
n2 = 250
sigma = sqrt(p1*(1-p1)/n1 + p2*(1-p2)/n2)
sigma

## [1] 0.03992794

• (a). 95% CI for \(\pi_{2} - \pi_{2}\): \(\mu \pm 1.96\sigma = 0.152 \pm 1.96(0.03992794)\) = [0.07374124, 0.2302588]

• (b). \(z = 3.806858 > z_{\alpha=0.01} = 2.326348\), hence reject the \(H_{0}\). Therefore, the warranty will increase the proportion of customers who will purchase a mower.

• (c). Offering warranty can significantly increase the proportion of people who will purchase a mower. However, whether the dealer should offer warranty or not depends on whether the profit from the additional sales of mower can cover the cost of offering warranty.

Problem 10.20

• (a).

  • 95% CI: \(\pi_{biofeedback} \pm 1.96\sigma = 0.56 \pm 1.96(0.01569713) = [0.5292336, 0.5907664]\)
    • \(\pi_{0} = 1240/2000 = 0.62\)
    • \(\pi_{biofeedback}\) = 560/1000 = 0.56
    • \(\sigma = \sqrt{\frac{\pi_{biofeedback}(1-\pi_{biofeedback})}{n}}\) = \(\sqrt{{0.56(1-0.56)/1000}}\) = 0.01569713
    • \(z = \frac{|\pi_{biofeedback} - \pi_{0}|}{\sigma}\) = \(\frac{|0.56 - 0.62|}{0.01569713}\) = 3.822355
  • 95% CI: \(\pi_{NSAID} \pm 1.96\sigma = 0.68 \pm 1.96(0.01475127) = [0.6510875, 0.7089125]\)
    • \(\pi_{0} = 1240/2000 = 0.62\)
    • \(\pi_{NSAID}\) = 680/1000 = 0.68
    • \(\sigma = \sqrt{\frac{\pi_{NSAID}(1-\pi_{NSAID})}{n}}\) = \(\sqrt{{0.68(1-0.68)/1000}}\) = 0.01475127
    • \(z = \frac{|\pi_{NSAID} - \pi_{0}|}{\sigma}\) = \(\frac{|0.68 - 0.62|}{0.01475127}\) = 4.067446

• (b). Yes. p-value = 4.204e-08 < 0.05. There is a strong evidence that the proportions of patients who experienced a significant reduction in pain are significantly different between the two treatments.

yes = c(560, 680)
no = c(440, 320)
pain_reduction = cbind(yes, no)
rownames(pain_reduction) = c("biofeedback", "NSAID")
chisq_test = chisq.test(pain_reduction)
chisq_test

## 
##  Pearson's Chi-squared test with Yates' continuity correction
## 
## data:  pain_reduction
## X-squared = 30.053, df = 1, p-value = 4.204e-08

• (c). 95% CI on the difference:

  • \(P_{biofeedback} - P_{NSAID}\) = 0.56 - 0.68 = -0.12
  • 95% CI on the difference: [-0.16321892, -0.07678108]

prop.test(pain_reduction)

## 
##  2-sample test for equality of proportions with continuity
##  correction
## 
## data:  pain_reduction
## X-squared = 30.053, df = 1, p-value = 4.204e-08
## alternative hypothesis: two.sided
## 95 percent confidence interval:
##  -0.16321892 -0.07678108
## sample estimates:
## prop 1 prop 2 
##   0.56   0.68

Problem 10.78

oneplus = c(10, 14, 19)
none = c(90, 86, 81)
tumors = data.frame(oneplus, none)
rownames(tumors) = c("control", "low_dose", "high_dose")
tumors

##           oneplus none
## control        10   90
## low_dose       14   86
## high_dose      19   81

• (a).
  • control: 10%
  • low dose: 14%
  • high dose: 19%
• (b). p-value = 0.1909 > 0.05. We fail to reject the \(H_{0}\). There is no strong evidence to show a significant difference in proportions among the three treatments.

chisq.test(tumors)

## 
##  Pearson's Chi-squared test
## 
## data:  tumors
## X-squared = 3.3119, df = 2, p-value = 0.1909

• (c). No. the Chi-square test failed to reject the null hypothesis.

Problem 2x2 contigency table

vote_for = c(11, 41)
vote_against = c(46, 2)
vote = cbind(vote_for, vote_against)
rownames(vote) = c("democrat", "republican")
vote

##            vote_for vote_against
## democrat         11           46
## republican       41            2

fisher.test(vote, conf.level = 0.99)

## 
##  Fisher's Exact Test for Count Data
## 
## data:  vote
## p-value = 2.029e-15
## alternative hypothesis: true odds ratio is not equal to 1
## 99 percent confidence interval:
##  0.0005700799 0.0834071130
## sample estimates:
## odds ratio 
## 0.01247519

prop.vote.fit = prop.test(vote)
prop.vote.fit

## 
##  2-sample test for equality of proportions with continuity
##  correction
## 
## data:  vote
## X-squared = 53.788, df = 1, p-value = 2.233e-13
## alternative hypothesis: two.sided
## 95 percent confidence interval:
##  -0.9011466 -0.6198652
## sample estimates:
##    prop 1    prop 2 
## 0.1929825 0.9534884