Answer to Question #152086 in Statistics and Probability for NURUL

Question

. Set up a 95% confidence interval estimate for the population mean, based on each of the
following sets of data, assuming that the population is normally distributed:
Set 1: 1, 1, 1, 1, 8, 8, 8, 8
Set 2: 1, 2, 3, 4, 5, 6, 7, 8
Explain why these data sets have different confidence intervals even though they have the
same mean and range.

Answer 1

The critical value for $\alpha=0.05$ is $z_c=z_{1-\alpha/2}=1.96$

The corresponding confidence interval is computed as shown below:

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

The confidence interval is equal to two margins of errors and a margin of error is equal to about 2 standard errors (for 95% confidence). A standard error is the standard deviation divided by the square root of the sample size.

The width of the confidence interval increases as the standard deviation increases.

Set 1: 1, 1, 1, 1, 8, 8, 8, 8

$$mean_1=\bar{x}_1$$

$$=\dfrac{1+1+1+1+8+8+8+8}{8}=4.5$$ $$Rank_1=8-1=7$$ $$\sigma_1^2=\dfrac{1}{8}((1-4.5)^2+(1-4.5)^2+(1-4.5)^2+(1-4.5)^2$$

$$+(8-4.5)^2+(8-4.5)^2+(8-4.5)^2+(8-4.5)^2)$$

$$=12.25$$

$$\sigma_1=\sqrt{12.25}=3.5$$

$$CI_1=(\bar{x}_1-z_c\times\dfrac{\sigma_1}{\sqrt{n_1}}, \bar{x}_1+z_c\times\dfrac{\sigma_1}{\sqrt{n_1}})$$

$$=(4.5-1.96\times\dfrac{3.5}{\sqrt{8}}, 4.5+1.96\times\dfrac{3.5}{\sqrt{8}})$$

$$=(2.075,6.925)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $2.075<\mu<6.925,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(2.075,6.925).$

Set 2: 1, 2, 3, 4, 5, 6, 7, 8

$$mean_2=\bar{x}_2$$

$$=\dfrac{1+2+3+4+5+6+7+8}{8}=4.5$$ $$Rank_2=8-1=7$$ $$\sigma_2^2=\dfrac{1}{8}((1-4.5)^2+(2-4.5)^2+(3-4.5)^2+(4-4.5)^2$$

$$+(5-4.5)^2+(6-4.5)^2+(7-4.5)^2+(8-4.5)^2)$$

$$=5.25$$

$$\sigma_2=\sqrt{5.25}\approx2.291288$$

$$CI_2=(\bar{x}_2-z_c\times\dfrac{\sigma_2}{\sqrt{n_2}}, \bar{x}_2+z_c\times\dfrac{\sigma_2}{\sqrt{n_2}})$$

$$=(4.5-1.96\times\dfrac{2.291288}{\sqrt{8}}, 4.5+1.96\times\dfrac{2.291288}{\sqrt{8}})$$

$$=(2.912,6.088)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $2.912<\mu<6.088,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(2.912,6.088).$