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Distribution of Sample Means

Distribution of Sample means: A sampling distribution of sample means is a distribution obtained by using the means computed from random samples of a specific size taken from a population.
Distribusi rata-rata sampel: Distribusi sampling dari rata-rata sampel adalah distribusi yang diperoleh dengan menggunakan rata-rata yang dihitung dari sampel acak dari ukuran tertentu yang diambil dari populasi.
Sampling error is the difference between the sample measure and the corresponding population measure due to the fact that the sample is not a perfect representation of the population.
Kesalahan sampling adalah perbedaan antara ukuran sampel dan ukuran populasi yang sesuai karena sampel tidak mewakili populasi dengan sempurna.

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Properties of the Distribution of Sample Means

The mean of the sample means will be the same as the population mean.
Rata-rata rata-rata sampel akan sama dengan rata-rata populasi.
The standard deviation of the sample means will be smaller than the standard deviation of the population, and it will be equal to the population standard deviation divided by the square root of the sample size.
Simpangan baku rata-rata sampel akan lebih kecil daripada simpangan baku populasi, dan akan sama dengan simpangan baku populasi dibagi dengan akar kuadrat dari ukuran sampel.

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Properties of the Distribution of Sample Means - Example

Suppose a professor gave an 8-point quiz to a small class of four students. The results of the quiz were 2, 6, 4, and 8. Assume the four students constitute the population.
The mean of the population is μ=2+6+4+84=5\mu = \dfrac{2+6+4+8}{4} = 5.
The standard deviation of the population is     σ=(2−5)2+(6−5)2+(4−5)2+(8−5)24=2.236~~~~~\sigma = \sqrt{\dfrac{(2-5)^2 + (6-5)^2 + (4-5)^2 + (8-5)^2}{4}} = 2.236.
The graph of the distribution of the scores is uniform.
Distribution of Scores
Next we will consider all samples of size 2 taken with replacement.
SampleMeanSampleMean
2,226,24
2,436,45
2,646,66
2,856,87
4,238,25
4,448,46
4,658,67
4,868,88

Frequency Distribution of the Sample Means
X-bar (mean)2345678
Frequency1234321

Graph of the Sample MeansSample Means

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Mean and Standard Deviation of the Sample Means

Mean of Sample Means
     μXˉ=∑Xn=2+3+⋯+816=8016=5~~~~~\mu_{\bar{X}} = \dfrac{\sum X}{n} = \dfrac{2+3+\cdots+8}{16} = \dfrac{80}{16} = 5.
which is the same as the population mean. Thus μ_Xˉ=μ\mu\_{\bar{X}} = \mu.
The standard deviation of the sample means is
     σXˉ=(2−5)2+(3−5)2+⋯+(8−5)216=2.2362=1.581~~~~~\sigma_{\bar{X}} = \sqrt{\dfrac{(2-5)^2 + (3-5)^2 + \cdots + (8-5)^2}{16}} = \dfrac{2.236}{\sqrt{2}} = 1.581.
This is the same as σ2\dfrac{\sigma}{\sqrt{2}}.
The Standard Error of the Mean
The standard deviation of the sample means is called the standard error of the mean. Hence σXˉ=σn\sigma_{\bar{X}} = \dfrac{\sigma}{\sqrt{n}}.

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The Central Limit Theorem

As the sample size n increases, the shape of the distribution of the sample means taken from a population with mean μ\mu and standard deviation of σ\sigma will approach a normal distribution. As previously shown, this distribution will have a mean μ\mu and standard deviation σn\dfrac{\sigma}{\sqrt{n}}.
Saat ukuran sampel n meningkat, bentuk distribusi rata-rata sampel yang diambil dari populasi dengan rata-rata μ\mu dan simpangan baku σ\sigma akan mendekati distribusi normal. Seperti yang telah ditunjukkan sebelumnya, distribusi ini akan memiliki rata-rata μ\mu dan simpangan baku σn\dfrac{\sigma}{\sqrt{n}}.
The central limit theorem can be used to answer questions about sample means in the same mannerthat the normal distribution can be used to answer questions about individual values. The only difference is that a new formula must be used for the z- values. It is z=Xˉ−μσnz = \dfrac{\bar{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}.
Teorema batas pusat dapat digunakan untuk menjawab pertanyaan tentang rata-rata sampel dengan cara yang sama seperti distribusi normal dapat digunakan untuk menjawab pertanyaan tentang nilai individu. Satu-satunya perbedaan adalah bahwa rumus baru harus digunakan untuk nilai z. Ini adalah z=Xˉ−μσnz = \dfrac{\bar{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}.

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The Central Limit Theorem - Example

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Question 1

A.C. Neilsen reported that children between the ages of 2 and 5 watch an average of 25 hours of TV per week. Assume the variable is normally distributed and the standard deviation is 3 hours. If 20 children between the ages of 2 and 5 are randomly selected
find the probability that the mean of the number of hours they watch TV is greater than 26.3 hours.
Solution :
  • The standard deviation of the sample means is σn=320=0.671\dfrac{\sigma}{\sqrt{n}} = \dfrac{3}{\sqrt{20}} = 0.671.
  • The z-value is z=Xˉ−μσn=26.3−250.671=1.94z = \dfrac{\bar{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}} = \dfrac{26.3-25}{0.671} = 1.94.
  • Thus P(z>1.94)=0.5−0.4732=0.0268P(z > 1.94) = 0.5 - 0.4732 = 0.0268.
That is, the probability of obtaining a sample mean greater than 26.3 is 0.0262 = 2.62%.
Question 1

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Question 2

The average age of a vehicle registered in the United States is 8 years, or 96 months. Assume the standard deviation is 16 months. If a random sample of 36 cars is selected
find the probability that the mean of their age is between 90 and 100 months.
Solution :
  • The standard deviation of the sample means is σn=1636=2.6667\dfrac{\sigma}{\sqrt{n}} = \dfrac{16}{\sqrt{36}} = 2.6667.
  • The two z-values are :
    •  ~
    •      z1=90−962.6667=−2.25~~~~~z_1 = \dfrac{90-96}{2.6667} = -2.25 . and
    •   ~~
    •      z2=100−962.6667=1.5~~~~~z_2 = \dfrac{100-96}{2.6667} = 1.5 .
    •   ~~
  • Thus P(−2.25<z<1.5)=0.4878+0.4332=0.921P(-2.25 < z < 1.5) = 0.4878 + 0.4332 = 0.921 or 92.1%.

Question 2