Email : ferdinand.widjaya@student.matanauniversity.ac.id
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15810.
For any particular random sample, we can always compute it’s sample mean. Although most often it is not the actual population mean, it does serve as a good point estimate. Note that, the population-mean is a measure of the center or “average” value in the whole population of a variable measured. Accordingly, the sample mean is a sample estimate of the population mean. It is the same measure of center, obtained from a sample. The variable in your sample must be measured at the interval or ratio level.
Find a point estimate of average university student Age with the sample data from survey!
library(MASS)
agesurvey = survey$Age # select the Age Column
mean(agesurvey, na.rm=TRUE) # the point estimate of student age## [1] 20.37451The Average Students Age is Around 20y.o
In some cases we have improve the quality of a sample survey by
increasing the sample size. We can use the following formula to provide
the sample size needed under the requirement of population mean interval
estimate at
(1−α) confidence level, margin of error E , and population variance σ^2
. Here, zα/2 is the 100(1−α/2) percentile of the standard normal
distribution.
Assume the population standard deviation
σ of the student height in survey is 9.48. Find the sample size needed
to achieve a 1.2 centimeters margin of error at 95% confidence
level.
zstar = qnorm(.975) # quantiles (95% confidence level)
sigma = 9.48 # assume population standard deviation
E = 1.2 # expected error
zstar^2*sigma^2/ E^2 # sampling size ## [1] 239.7454Based on the assumption of population standard deviation being 9.48, it needs a sample size of 240 to achieve a 1.2 centimeters margin of error at 95% confidence level.
Improve the quality of a sample survey by increasing the sample size with unknown standard deviation σ !.
zstar = qnorm(.975) # quantiles (95% confidence level)
sigma = 9.48 # assume population standard deviation
E = 0.72 # Reduce the Expected Error by 40%
zstar^2*sigma^2/ E^2 # sampling size ## [1] 665.9596Dengan asumsi bahwa SD = 9.48 MaKa buth sampel sebanyak 666 untuk mendapatkan nilai margin error 0.72cm ddengan CI 95%
The quality of a sample survey can be improved by increasing the sample size. The formula below provide the sample size needed under the requirement of population proportion interval estimate at (1−α) confidence level, margin of error , and planned proportion estimate p . Here, zα/2 is the 100(1−α/2) percentile of the standard normal distribution. n=(zα/2)^2 p(1−p) / E^2
Using a 50% planned proportion estimate, find the sample size needed to achieve 5% margin of error for the female student survey at 95% confidence level.
zstar = qnorm(.975) # quantiles (95% confidence level)
p = 0.5 # 50% planned proportion estimate
E = 0.05 # expected error
zstar^2*p*(1-p)/E^2 # sampling size## [1] 384.1459Dengan Proporasi 50-50, untuk mendapatkan error rate 5% maka sibtuhkan 385 Siswa Perempuan.
Assume you don’t have planned proportion estimate, find the sample size needed to achieve 5% margin of error for the male student survey at 95% confidence level!
zstar = qnorm(.975) # quantiles (95% confidence level)
p = 0.75 # Unplanned proportion estimate, Proportion We Received From Data is Male 75-25 Female
E = 0.05 # expected error
zstar^2*p*(1-p)/E^2 # sampling size## [1] 288.1094Dengan Proporsi Laki-Laki ke Perempuan 75-25 sebagai contoh, dibutuhkan sampel laki-laki sebanyak 289 orang.
Perform confidence intervals analysis on this dataset from 2004 that includes data on average hourly earnings, marital status, gender, and age for thousands of people.
cps = read.csv('cps04.csv')mean(cps$age, na.rm=T)## [1] 29.75445Rata-Rata berusia 29-30 an Tahun
SE = (sd(cps$age))/ 7986
E = qt(.975, 7985)*SE
E## [1] 0.000709662mean(cps$age, na.rm=T) + c(-E, E) # confidence interval as told## [1] 29.75374 29.75515Dengan Confidence Interval 95% kita bisa tahu bahwa sebaran atau erornya sangat kecil, jadi kebanyakan persebaran data ada di usia 29-30
t.test(cps$age)##
## One Sample t-test
##
## data: cps$age
## t = 919.71, df = 7985, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
## 29.69103 29.81786
## sample estimates:
## mean of x
## 29.75445single = cps$bachelor
a = sum(cps$bachelor == "1")
b = length(single)
propmar = a/b
propmar## [1] 0.4557976Estimasi Sebanyak 45,57% dari proporsi masih Single atau Belum Menikah.
prop.test(a,b)##
## 1-sample proportions test with continuity correction
##
## data: a out of b, null probability 0.5
## X-squared = 62.237, df = 1, p-value = 3.045e-15
## alternative hypothesis: true p is not equal to 0.5
## 95 percent confidence interval:
## 0.4448359 0.4668022
## sample estimates:
## p
## 0.4557976Dengan Confidence Interval, kita bisa memperkirakan pada populasi bahwa sekitar 44.48% sampai 46.68% Masih Single.
female = cps$female
aa = sum(cps$female == "1")
bb = length(female)
propfe = aa/bb
propfe## [1] 0.414851Estimasi bahwa 41.48% Populasi adalah WAnita
prop.test(aa,bb)##
## 1-sample proportions test with continuity correction
##
## data: aa out of bb, null probability 0.5
## X-squared = 231.26, df = 1, p-value < 2.2e-16
## alternative hypothesis: true p is not equal to 0.5
## 95 percent confidence interval:
## 0.4040262 0.4257582
## sample estimates:
## p
## 0.414851Dengan Confidence Interval pada 95% diperkirakan pada populasi, 40.40% sampai 42.57% adalah WAnita.