Answer to Question #207310 in Statistics and Probability for juanid

Question

Estimation and Confidence Intervals

The Greater Pittsburgh Area Chamber of Commerce wants to estimate the mean time workers who are employed in the downtown area spend getting to work. A sample of 15 workers reveals the following number of minutes spent traveling.

29, 38, 38, 33, 38, 21, 45, 34, 40, 37, 37, 42, 30, 29, 35

Requirements:

i) Develop a 90%, 95% and 98 percent confidence interval for the population mean?

ii) Compare and interpret the result?

Answer 1

$$29, 38, 38, 33, 38, 21, 45, 34, 40, 37, 37, 42, 30, 29, 35$$

$n=15$

Sample mean

$$\bar{x}= \frac{\displaystyle\sum_{i=1}^nx_i}{n} =\frac{1}{15} (29+38+38+33+38+21$$

$$+45+34+40+37+37+42+30+29+35)$$

$$=\dfrac{526}{15}\approx35.066667$$

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

$$=\frac{1}{15-1}((29-\dfrac{526}{15})^2+(38-\dfrac{526}{15})^2$$

$$+(38-\dfrac{526}{15})^2+(33-\dfrac{526}{15})^2+(38-\dfrac{526}{15})^2$$

$$+(21-\dfrac{526}{15})^2+(45-\dfrac{526}{15})^2+(34-\dfrac{526}{15})^2$$

$$+(40-\dfrac{526}{15})^2+(37-\dfrac{526}{15})^2+(37-\dfrac{526}{15})^2$$

$$+(42-\dfrac{526}{15})^2+(38-\dfrac{526}{15})^2+(29-\dfrac{526}{15})^2$$

$$+(35-\dfrac{526}{15})^2\approx36.2095238$$

$$s=\sqrt{s^2}\approx6.017435$$

i)

90%

The critical value for $\alpha=0.1$ and $df=n-1=15-1=14$ degrees of freedom is $t_c=1.76131.$ The corresponding confidence interval is computed as shown below:

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

$$=(35.066667-1.76131\times \dfrac{6.017435}{\sqrt{15}},$$

$$35.066667+1.76131\times \dfrac{6.017435}{\sqrt{15}} )$$

$$\approx(32.33013, 37.80321)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $32.33013<\mu<37.80321,$ which indicates that we are 90% confident that the true population mean $\mu$ is contained by the interval $(32.33013, 37.80321).$

95%

The critical value for $\alpha=0.05$ and $df=n-1=15-1=14$ degrees of freedom is $t_c= 2.144787.$ The corresponding confidence interval is computed as shown below:

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

$$=(35.066667- 2.144787\times \dfrac{6.017435}{\sqrt{15}},$$

$$35.066667+ 2.144787\times \dfrac{6.017435}{\sqrt{15}} )$$

$$\approx(31.73432, 38.39901)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $31.73432<\mu<38.39901,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(31.73432, 38.39901).$

98%

The critical value for $\alpha=0.02$ and $df=n-1=15-1=14$ degrees of freedom is $t_c= 2.624494.$ The corresponding confidence interval is computed as shown below:

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

$$=(35.066667- 2.624494\times \dfrac{6.017435}{\sqrt{15}},$$

$$35.066667+ 2.624494\times \dfrac{6.017435}{\sqrt{15}} )$$

$$\approx(30.98900, 39.14433)$$

Therefore, based on the data provided, the 98% confidence interval for the population mean is $30.98900<\mu<39.14433,$ which indicates that we are 98% confident that the true population mean $\mu$ is contained by the interval $(30.98900, 39.14433).$

ii)

The 90% confidence interval is $(32.33013, 37.80321).$

The 95% confidence interval is $(31.73432, 38.39901).$

The 98% confidence interval is $(30.98900, 39.14433).$

The 95% confidence interval is wider than the 90% confidence interval.

The 98% confidence interval is wider than the 95% confidence interval and wider than the 90% confidence interval.

Increasing the confidence level increases the error bound, making the confidence interval wider.

Decreasing the confidence level decreases the error bound, making the confidence interval narrower.