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.