91影视

The results of the testGrade-Point Averageare to be standardized on a scale between 0 and 4. Using the range rule of thumb, the estimated value ofis 1.The confidence level is 95% and it implies that the sample mean is within the 0.01 Grade-Point of the true mean.

The sample size n for estimating the true population mean can be determined by using the following formula.

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

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 95% corresponds to $伪=0.05鈥�鈥�鈥�鈥�\text{and}鈥�鈥�鈥�鈥�\frac{伪}{2}=0.025$.

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

Mathematically,

$$\begin{gathered} P\left(z<z_{\frac{伪}{2}}\right)=1-\frac{伪}{2} \\ =0.975 \end{gathered}$$

From the standard normal table, the area of 0.99 is observed corresponding to the row value 1.9 and column value 0.06, which implies $z_{\frac{伪}{2}}$is 1.96.

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

$$\begin{gathered} \text{n}={\left[\frac{z_{\frac{伪}{2}}脳蟽}{\text{E}}\right]}^2 \\ ={\left[\frac{1.96脳1}{0.01}\right]}^2 \\ =38416 \end{gathered}$$

Thus, with 38416 samples values, you can be95% confident that your sample mean lies within 0.01 Grade-points of the true mean.

The sample size required to estimate the Mean Grade-Point Averageis 38416. The estimated sample size is too large; so the sample size does not seem practical.