Prueba de hipótesis para una media
| \(Ho\) : \(\mu = \mu_o\) |
\(Ho\) : \(\mu \leq \mu_o\) |
\(Ho\) : \(\mu \geq \mu_o\) |
| \(Ha\) : \(\mu \neq \mu_o\) |
\(Ha\) : \(\mu > \mu_o\) |
\(Ha\) : \(\mu < \mu_o\) |
Supuestos:
| X normal |
| Varianza desconocida |
| \(Ho\) : \(\mu \geq 5\) |
| \(Ha\) : \(\mu < 5\) |
#-------------------------------------------------------------------------------
# Problema 2
t=c(4.21,5.55,3.02,5.13,4.77,2.34,5.42,4.50,6.10,3.80,5.12,6.46,6.19,3.79,3.54)
mean(t)
[1] 4.662667
sd(t)
[1] 1.210658
t.test(t,mu=5, alternative="less")
One Sample t-test
data: t
t = -1.0792, df = 14, p-value = 0.1494
alternative hypothesis: true mean is less than 5
95 percent confidence interval:
-Inf 5.213235
sample estimates:
mean of x
4.662667
| \(Ho\) : \(p \geq 5\) |
| \(Ha\) : \(p < 5\) |
#-------------------------------------------------------------------------------
#Problema 3
z=(24/40-.76)/(sqrt(.76*(1-.76)/40))
prop.test(24,40,0.76,alternative="less")
1-sample proportions test with continuity correction
data: 24 out of 40, null probability 0.76
X-squared = 4.7711, df = 1, p-value = 0.01447
alternative hypothesis: true p is less than 0.76
95 percent confidence interval:
0.0000000 0.7282033
sample estimates:
p
0.6
| \(Ho\) : \(\mu_1 \geq \mu_2\) |
| \(Ha\) : \(\mu_1 < \mu_2\) |
#---------------------------------------------------------------------------------
# Problema 4
n1=36 ; mx1=6 ; sx1=4
n2=40 ; mx2=8.2; sx2=4.3
F=sx1^2/sx2^2
RdeRF=qf(c(0.025,0.975),35,39)
#t.test(x1,x2,mu=0, alternative = "less")
s2p=((n1-1)*sx1^2+(n2-1)*sx2^2)/(n1+n2-2)
sp=sqrt(s2p)
T4=(mx1-mx2)/(sp*sqrt(1/n1+1/n2))
RdeRT4=qt(0.05,(n1+n2-2))
| \(Ho\) : \(\mu_{g1} \geq \mu_{g2}\) |
| \(Ha\) : \(\mu_{g1} < \mu_{g2}\) |
#-----------------------------------------------------------------------------------
# Problema 5
p5=3/40
z5=(p5-0.05)/sqrt(0.05*0.95/40)
#----------------------------------------------------------------------------------
# Problema 6
g1=c(75,76,74,80,72,798,76,73,72,75)
g2=c(86,78,86,84,81,79,78,84,88,80)
mean(g1);sd(g1)
[1] 147.1
[1] 228.715
mean(g2);sd(g2)
[1] 82.4
[1] 3.657564
var.test(g1,g2)
F test to compare two variances
data: g1 and g2
F = 3910.3, num df = 9, denom df = 9, p-value = 8.882e-15
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
971.2524 15742.6704
sample estimates:
ratio of variances
3910.257
t.test(g1,g2)
Welch Two Sample t-test
data: g1 and g2
t = 0.89445, df = 9.0046, p-value = 0.3944
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-98.92101 228.32101
sample estimates:
mean of x mean of y
147.1 82.4
| \(Ho\) : \(p_1 = p_2\) |
| \(Ha\) : \(p \neq p_2\) |
#---------------------------------------------------------------------------------
# Problema 7
n1=400 ;x1=80
n2=400 ; x2=88
prop.test(c(80,88),c(400,400))
2-sample test for equality of proportions with continuity correction
data: c(80, 88) out of c(400, 400)
X-squared = 0.3692, df = 1, p-value = 0.5434
alternative hypothesis: two.sided
95 percent confidence interval:
-0.07893199 0.03893199
sample estimates:
prop 1 prop 2
0.20 0.22
| \(Ho\) : \(\mu_1 \geq \mu_2\) |
| \(Ha\) : \(\mu_1 < \mu_2\) |
#-----------------------------------------------------------------------------------
# Problema 8
x1=c(45,73,46,124,30,57,83,34,26,17)
x2=c(36,60,44,119,35,51,77,29,24,11)
d=x1-x2
mean(d)
[1] 4.9
sd(d)
[1] 4.72464
t.test(x1,x2,paired = TRUE)
Paired t-test
data: x1 and x2
t = 3.2796, df = 9, p-value = 0.009535
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
1.520196 8.279804
sample estimates:
mean of the differences
4.9
| \(Ho\) : \(\mu_{pa} \geq \mu_{pd}\) |
| \(Ha\) : \(\mu_{pa} < \mu_{pd}\) |
#------------------------------------------------------------------------------------
#Problema 9
pa=c(104.5,89,84.5,106,90,96,79,90,85,76.5,91.5,82.5,100.5,89.5,121.5,72)
pd=c(98,85.5,85,103.5,88.5,95,79.5,90,82,76,89.5,81,99.5,86.5,115.5,70)
d=pa-pd
mean(d)
[1] 2.0625
sd(d)
[1] 2.032035
t.test(pa,pd,paired = TRUE)
Paired t-test
data: pa and pd
t = 4.06, df = 15, p-value = 0.001026
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.9797049 3.1452951
sample estimates:
mean of the differences
2.0625
| \(Ho\) : \(X_{g1} \sim norm\) |
\(Ho\) : \(\sigma^{2}_{g1} = \sigma^{2}_{g2}\) |
\(Ho\) : \(\mu_{g1} \geq \mu_{g2}\) |
| \(Ha\) : \(X_{g2} no \sim norm\) |
\(Ha\) : \(\sigma^{2}_{g1} \neq \sigma^{2}_{g2}\) |
\(Ha\) : \(\mu_{g1} < \mu_{g2}\) |
#-----------------------------------------------------------------------
# Problema 10
g1=c(37,19,21,35,16,4,0,12,63,25,12,15)
g2=c(24,42,18,15,0,9,10,20,22,13)
mean(g1);sd(g1)
[1] 21.58333
[1] 17.01581
mean(g2);sd(g2)
[1] 17.3
[1] 11.20565
var.test(g1,g2)
F test to compare two variances
data: g1 and g2
F = 2.3058, num df = 11, denom df = 9, p-value = 0.22
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.5894187 8.2731556
sample estimates:
ratio of variances
2.30585
t.test(g1,g2)
Welch Two Sample t-test
data: g1 and g2
t = 0.70719, df = 19.104, p-value = 0.488
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-8.389007 16.955673
sample estimates:
mean of x mean of y
21.58333 17.30000
