Question

From our previous exam in Statistics, sample scores from a normal population were taken for evaluation purposes of how the students responded the test. The scores are 38, 27, 34, 50, 33, 36, 21, 27, 22, 29, 27, 27, 22, 29, 31, 28 asssume that the score are normally distributed

Find the value of the point estimate for the population.

Compute for the margin of error at 99% confidence level.

Find 99% confidence interval for the population mean.

Interpret the results.

Answer 1

1.

$$\bar{x}=\dfrac{1}{16}(38+27+ 34+50+33+36+21+27+22$$

$$+29+27+27+22+29+31+28)=30.0625$$

2.

$$s^2=\dfrac{1}{n-1}\displaystyle\sum_{i=1}^n(x_i-\bar{x})^2$$

$$s^2\approx51.795833$$

$$s=\sqrt{s^2}\approx7.196932$$

The critical value for $\alpha=0.01$ and $df=n-1=15$ degrees of freedom is $t_c=2.946712.$

The margin of error is

$$ME=t_c\times\dfrac{s}{\sqrt{n}}=2.946712\times\dfrac{7.196932}{\sqrt{16}}$$

$$\approx5.301821$$

3. The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$\approx(30.0625-2.946712\times\dfrac{7.196932}{\sqrt{16}},$$

$$30.0625-2.946712\times\dfrac{7.196932}{\sqrt{16}})$$

$$\approx(24.7607, 35.3642)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $24.7607<\mu<35.3642,$ which indicates that we are 99% confident that the true population mean $\mu$ is contained by the interval $(24.7607, 35.3642).$