Question

A journalist would like to report on the traffic–related tardiness incurred by employees

in a certain city per year. To do this, he asked 25 employees and found out that they were

late for an average of 8.5 times a year due to traffic. The sample standard deviation was found to be in 1.5 times.

1. At a confidence level of 90 percent, determine the confidence interval that will

estimate the population mean.

2. At a confidence level of 95 percent, determine the confidence interval that will

estimate the average number of traffic-related tardiness per year by employees in

a city.

3. Which between the answers to items a and b gave a wider margin of error? What

does this say about the relationship between the confidence level and the margin

of error?

Answer 1

1.

The critical value for $\alpha=0.10,$ and $df=n-1=25-1=24$ degrees of freedom is $t_c=1.710882.$

The corresponding confidence interval is computed as shown below:

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

$$=(8.5-1.710882\times\dfrac{1.5}{\sqrt{25}}, 8.5+1.710882\times\dfrac{1.5}{\sqrt{25}})$$

$$=(7.987, 9.013)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $7.987<\mu<9.013,$ which indicates that we are 90% confident that the true population mean $\mu$ is contained by the interval $(7.987, 9.013).$

2.

The critical value for $\alpha=0.05,$ and $df=n-1=25-1=24$ degrees of freedom is $t_c=2.063899.$

The corresponding confidence interval is computed as shown below:

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

$$=(8.5-2.063899\times\dfrac{1.5}{\sqrt{25}}, 8.5+2.063899\times\dfrac{1.5}{\sqrt{25}})$$

$$=(7.881, 9.119)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $7.881<\mu<9.119,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(7.881, 9.119).$

3.

The 90% confidence interval for the population mean is $7.987<\mu<9.013.$

The 95% confidence interval for the population mean is $7.881<\mu<9.119.$

The higher the confidence level, the wider the confidence interval is (if everything else is equal).

As the confidence level increases, the margin of error increases.