91影视

Theweights of 153 randomly selected adult males, has a standard deviation of 17.65 kg. Assume thattheweights of the male Statistics students have less variation than the weights of the population of adult males.

It is required to be 90% confident that sample mean is within 1.5 kg of the true mean..

The sample size n needs to be determined to estimate the value of the population mean $渭$.

The sample size n can be determined by using the following formula.

$n={\left[\frac{z_{\frac{伪}{2}}脳蟽}{E}\right]}^2鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�...\left(1\right)$

Here, E is the margin of error.

The$z_{\frac{伪}{2}}$is a z score that separates an area of $\frac{伪}{2}$in the right tail of the standard normal distribution.

The confidence level of 90% corresponds to $伪=0.10鈥�鈥�鈥�鈥�\text{and}鈥�鈥�鈥�鈥�\frac{伪}{2}=0.05$.

The value$z_{\frac{伪}{2}}$, has the cumulative area of $1-\frac{伪}{2}$ to its left .

Mathematically,

$$\begin{gathered} P\left(z<z_{\frac{伪}{2}}\right)=1-\frac{伪}{2} \\ =0.95 \end{gathered}$$

From the standard normal table, the area of 0.95 is observed corresponding to the row value 1.6 and between column value 0.04 and column value 0.05, which implies $z_{\frac{伪}{2}}$ is 1.645.

The sample size is calculated by substituting the values of $z_{\frac{伪}{2}}$,$蟽$, and E in equation (1).

$$\begin{gathered} \text{n}={\left[\frac{z_{\frac{伪}{2}}脳蟽}{\text{E}}\right]}^2 \\ ={\left[\frac{1.645脳17.65}{1.5}\right]}^2 \\ =374.66 \\ =375鈥�鈥�鈥�鈥�\left(\text{rounded}鈥�鈥�\text{off}\right) \end{gathered}$$

Thus, with 375 samples values, you can be90% confident that your sample mean lies within 1.5kg of the true mean.

The sample size required to estimate the Mean Weight of Male Statistics Students is 374.

Yes,itdoes seem reasonable to assume that the weights of male Statistics students have less variation than the weights of the population of adult males as adult males tend show more dispersed weight measures as compared to the students.