Respuesta :
Answer:
(a) The variance decreases.
(b) The variance increases.
Step-by-step explanation:
According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.
Then, the mean of the sample mean is given by,
[tex]\mu_{\bar x}=\mu[/tex]
And the standard deviation of the sample mean is given by,
[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}[/tex]
The standard deviation of sample mean is inversely proportional to the sample size, n.
So, if n increases then the standard deviation will decrease and vice-versa.
(a)
The sample size is increased from 64 to 196.
As mentioned above, if the sample size is increased then the standard deviation will decrease.
So, on increasing the value of n from 64 to 196, the standard deviation of the sample mean will decrease.
The standard deviation of the sample mean for n = 64 is:
[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{64}}=0.7[/tex]
The standard deviation of the sample mean for n = 196 is:
[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{196}}=0.4[/tex]
The standard deviation of the sample mean decreased from 0.7 to 0.4 when n is increased from 64 to 196.
Hence, the variance also decreases.
(b)
If the sample size is decreased then the standard deviation will increase.
So, on decreasing the value of n from 784 to 49, the standard deviation of the sample mean will increase.
The standard deviation of the sample mean for n = 784 is:
[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{784}}=0.2[/tex]
The standard deviation of the sample mean for n = 49 is:
[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{49}}=0.8[/tex]
The standard deviation of the sample mean increased from 0.2 to 0.8 when n is decreased from 784 to 49.
Hence, the variance also increases.
Variance has decreased in first case and Variance has increased in second case.
Variation based problem:
What information do we have?
Mean tensile stength = 78.3 kilograms
Standard variance = 5.6 kilogram
Variance = sigma² / n
A.
Variance of sample mean with sample size 64 = 5.6² / 64
Variance of sample mean with sample size 64 = 0.49
Variance of sample mean with sample size 196 = 5.6² / 196
Variance of sample mean with sample size 196 = 0.16
Variance has decreased.
B.
Variance of sample mean with sample size 784 = 5.6² / 784
Variance of sample mean with sample size 784 = 0.04
Variance of sample mean with sample size 49 = 5.6² / 49
Variance of sample mean with sample size 49 = 0.64
Variance has increased.
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