Question

Student Height (x) 1.47 Hazel Kim Mee Ann Mae 1.58 Grace Ann 1.67 Jethro 1.69 1.51 Chelsea Mae TOTAL 7.92 STEPS: 1. Solve for the mean of the population u. 2. Solve for the mean of the sampling distribution of the sample means 3. Compare p and 4. Solve for the variance (0 ?) and the standard deviation (a) of the population. 5. Solve the variance (0 x) and the standard deviation (ox) of the sampling distribution of the sample means 6. Compare o and on

Answer 1

 (1)The population mean is calculated with the following formula below.

$\mu=\dfrac{1}{N}\sum X_i=\dfrac{1.47+1.58+1.67+1.69+1.51}{5}=\dfrac{7.92}{5}=1.584$

(2)The total number of sample can be drawn from above population $^NC_n=^5C_2=10\ samples$

The sample of two of the student chosen at random from the above population of size 5 be

The sampling distribution of mean is calculated as below

$\mu_{\bar X}=\dfrac{1}{^NC_n}\sum \bar X_i=\dfrac{15.84}{10}=1.584$

(3)The difference is given below

$\mu-\mu_{\bar x}=0$

Therefore sample mean is an unbiased estimator of population mean.

(4) The population variance is calculated with the following formula below:

$\sigma^2=\dfrac{1}{N}(\sum X^2_{\bar x})-\mu^2\\\Rightarrow\sigma^2=\frac{1}{5}(12.58)-(1.584)^2\\\Rightarrow \sigma^2=0.0074$

Population Standard Deviation is equal to

$\Rightarrow \sigma=\sqrt{\sigma^2}=\sqrt{0.0074}=0.086$

(5)  The sample variance of sampling distribution of sampling mean is calculated below

$\sigma^2_{\bar X}=\dfrac{1}{^NC_n}\sum (\bar X_i-\mu_{\bar X})^2$

The calculations are given below.

$\sigma^2_{\bar X}=\dfrac{1}{^NC_n}\sum (\bar X_i-\mu_{\bar X})^2=\dfrac{0.02784}{10}=0.002784$

population standard deviation is equal to

$\sigma_{\bar X}=\sqrt{\sigma^2_{\bar X}}=\sqrt{0.002784}=0.05276$

(6) The comparison is given below as under

$\sigma-\sigma_{\bar X}=0.086-0.05276=0.03324$

The Sample standard deviation is not unbiased estimator for Population Standard Deviation.