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 <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 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,
![\mu_{\bar x}=\mu](https://tex.z-dn.net/?f=%5Cmu_%7B%5Cbar%20x%7D%3D%5Cmu)
And the standard deviation of the sample mean is given by,
![\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Csigma_%7B%5Cbar%20x%7D%3D%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
The standard deviation of sample mean is inversely proportional to the sample size, <em>n</em>.
So, if <em>n</em> 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 <em>n</em> from 64 to 196, the standard deviation of the sample mean will decrease.
The standard deviation of the sample mean for <em>n</em> = 64 is:
![\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{64}}=0.7](https://tex.z-dn.net/?f=%5Csigma_%7B%5Cbar%20x%7D%3D%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%3D%5Cfrac%7B5.6%7D%7B%5Csqrt%7B64%7D%7D%3D0.7)
The standard deviation of the sample mean for <em>n</em> = 196 is:
![\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{196}}=0.4](https://tex.z-dn.net/?f=%5Csigma_%7B%5Cbar%20x%7D%3D%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%3D%5Cfrac%7B5.6%7D%7B%5Csqrt%7B196%7D%7D%3D0.4)
The standard deviation of the sample mean decreased from 0.7 to 0.4 when <em>n</em> 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 <em>n</em> from 784 to 49, the standard deviation of the sample mean will increase.
The standard deviation of the sample mean for <em>n</em> = 784 is:
![\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{784}}=0.2](https://tex.z-dn.net/?f=%5Csigma_%7B%5Cbar%20x%7D%3D%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%3D%5Cfrac%7B5.6%7D%7B%5Csqrt%7B784%7D%7D%3D0.2)
The standard deviation of the sample mean for <em>n</em> = 49 is:
![\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{49}}=0.8](https://tex.z-dn.net/?f=%5Csigma_%7B%5Cbar%20x%7D%3D%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%3D%5Cfrac%7B5.6%7D%7B%5Csqrt%7B49%7D%7D%3D0.8)
The standard deviation of the sample mean increased from 0.2 to 0.8 when <em>n</em> is decreased from 784 to 49.
Hence, the variance also increases.