EstatÃstica II - Exemplos
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CapÃtulo 5
Exemplo 5.2
- P([`x]B £ 167|m = 171) = P(Z £ -2.31) =
> pnorm(-2.31)
[1] 0.01044408
- b = P(Não Rejeitar H0 - H0 é Falsa)=
> 1 - dnorm(-0.4618802)
[1] 0.6414206
Exemplo 5.3
> dados <- c(498.8, 503.1, 497.6, 491.6, 499.3, 491.3, 499.8, 492.1,
+ 498.1, 493.2, 487.2, 489.8, 495.8, 498.2, 498.8, 485.7)
> t.test(dados, mu = 500, conf = 0.99, alternative = "less")
One Sample t-test
data: dados
t = -3.9825, df = 15, p-value = 0.0006007
alternative hypothesis: true mean is less than 500
99 percent confidence interval:
-Inf 498.2761
sample estimates:
mean of x
495.025
Exemplo 5.4
> xa <- c(3.4, 2.99, 3.21, 3.07, 3.01, 3.27, 3.23, 3.02)
> xa
[1] 3.40 2.99 3.21 3.07 3.01 3.27 3.23 3.02
> xb <- c(2.82, 3.16, 2.98, 3.04, 3.15, 3.2, 3, 3.01, 3.08, 3.06)
> xb
[1] 2.82 3.16 2.98 3.04 3.15 3.20 3.00 3.01 3.08 3.06
- Teste para as variâncias.
> var.test(xa, xb, conf.level = 0.95, alternative = "greater")
F test to compare two variances
data: xa and xb
F = 1.8569, num df = 7, denom df = 9, p-value = 0.1905
alternative hypothesis: true ratio of variances is greater than 1
95 percent confidence interval:
0.5639297 Inf
sample estimates:
ratio of variances
1.856877
- Teste para as médias.
> t.test(xa, xb, conf = 0.95, var.equal = T, alternative = "two.sided")
Two Sample t-test
data: xa and xb
t = 1.6443, df = 16, p-value = 0.1196
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.02892157 0.22892157
sample estimates:
mean of x mean of y
3.15 3.05
Exemplo 5.5
- Resistência de dois tipos de concreto.
> conc1 <- c(101.2, 102, 100.8, 102.3, 101.6)
> conc1
[1] 101.2 102.0 100.8 102.3 101.6
> conc2 <- c(100, 102.8, 101.5, 99, 102)
> conc2
[1] 100.0 102.8 101.5 99.0 102.0
- Teste para as variâncias.
> var.test(conc2, conc1)
F test to compare two variances
data: conc2 and conc1
F = 6.5414, num df = 4, denom df = 4, p-value = 0.09617
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.6810783 62.8274220
sample estimates:
ratio of variances
6.541436
- Teste das médias.
> t.test(conc1, conc2, alternative = "greater")
Welch Two Sample t-test
data: conc1 and conc2
t = 0.7037, df = 5.195, p-value = 0.2559
alternative hypothesis: true difference in means is greater than 0
95 percent confidence interval:
-0.9566956 Inf
sample estimates:
mean of x mean of y
101.58 101.06
Exemplo 5.6
> antes <- c(179, 200, 161, 170, 181, 190, 202, 220, 195, 165)
> depois <- c(160, 180, 161, 180, 165, 170, 196, 216, 170, 160)
> d <- mean(depois - antes)
> d
[1] -10.5
> t.test(antes, depois, paired = TRUE, conf = 0.99)
Paired t-test
data: antes and depois
t = 2.9811, df = 9, p-value = 0.01542
alternative hypothesis: true difference in means is not equal to 0
99 percent confidence interval:
-0.9464151 21.9464151
sample estimates:
mean of the differences
10.5
Exemplo 5.7
> require(corpora)
> prop.cint(98, 200, method = "z.score", correct = FALSE, conf.level = 0.95,
+ alternative = "less")
lower upper
1 0 0.547887
> z.score(98, 200, p = 0.57, correct = FALSE)
[1] -2.285248
> z.score.pval(98, 200, p = 0.57, correct = FALSE, alternative = "less")
[1] 0.01114915
Exemplo 5.8
- 2.
> ztab <- qnorm(0.99)
> ztab
[1] 2.326348
> kr <- 171
> nr <- 180
> kc <- 171
> nc <- 190
> pr <- kr/nr
> pc <- kc/nc
> pr
[1] 0.95
> pc
[1] 0.9
> p <- (kr + kc)/(nr + nc)
> p
[1] 0.9243243
- 3.
> zcalc <- ((pr - pc) - (0))/sqrt((p * (1 - p)) * ((1/nr) + (1/nc)))
> zcalc
[1] 1.817573
> p.valor <- 1 - pnorm(zcalc)
> p.valor
[1] 0.03456471
A função prop.test obtém o mesmo p-valor mas, considerando a distribuição c2
> prop.test(c(171, 171), c(180, 190), corre = FALSE, conf = 0.99,
+ alt = "g")
2-sample test for equality of proportions without continuity
correction
data: c(171, 171) out of c(180, 190)
X-squared = 3.3036, df = 1, p-value = 0.03456
alternative hypothesis: greater
99 percent confidence interval:
-0.01317965 1.00000000
sample estimates:
prop 1 prop 2
0.95 0.90
Observe que a estatÃstica de teste é baseada no teste de c2.
Exemplo 5.9
- 2.
> xq <- qchisq(0.975, 6)
> xq
[1] 14.44938
- 3.
> dados <- matrix(c(30, 18, 26, 13, 7, 8, 17), nrow = 1)
> chisq.test(dados)
Chi-squared test for given probabilities
data: dados
X-squared = 26.3529, df = 6, p-value = 0.0001913
Exemplo 5.10
- Renda e número de filhos por famÃlia em uma cidade.
> zero <- c(15, 8, 25)
> um <- c(27, 13, 30)
> dois <- c(50, 9, 12)
> mais <- c(43, 10, 8)
> tab5.13 <- data.frame(zero, um, dois, mais)
> # O próximo passo é apenas para que a tabela fique mais apresentável. Mas pode-se rodar o teste mesmo sem alterar o nome das linhas.
> dimnames(tab5.13)[[1]] <- c("< 2000", "2000 a 5000", "> 5000")
> tab5.13
zero um dois mais
< 2000 15 27 50 43
2000 a 5000 8 13 9 10
> 5000 25 30 12 8
> chisq.test(tab5.13)
Pearson's Chi-squared test
data: tab5.13
X-squared = 36.6212, df = 6, p-value = 2.087e-06
- P( c26 ³ 36,62) =
> p.valor <- round(1 - pchisq(36.6212, 6), 8)
> p.valor
[1] 2.09e-06
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On 26 May 2009, 15:57.