Question

Collect Sample data and use the methods of chapter 7 to construct confidence interval estimates of the population parameters. For example the mean number of hours that the students at your college work per week. You must include your data as well as the work and explanation. Give two sets of examples along with their respective data.

Answer 1

Example 1

Here, we want to find the confidence interval for the population mean time to reach college from the home or the place at where student lives. We randomly select the 8 students from the respective college and ask them about the time required to reach the college. The collected responses in minutes are given as below:

Time (Minutes)
22
21
28
23
20
28
30
29

We assume that above sample is coming from the normally distributed population. Suppose we want to find 95% confidence interval for the population mean time to reach college.

For the above sample data, we have

Sample mean = Xbar = 24.5

Sample standard deviation = S = 7.41

Sample size = n = 8

Degrees of freedom = df = n – 1 = 7

Confidence level = 95%

Critical t value = 2.3646

(by using t-table)

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 24.5 ± 2.3646*7.41/sqrt(8)

Confidence interval = 24.5 ± 2.3646*2.619830624

Confidence interval = 24.5 ± 6.1949

Lower limit = 24.5 - 6.1949 = 18.3051

Upper limit = 24.5 + 6.1949 = 30.6949

We are 95% confident that the population mean time to reach the college will be lies within 18.31 minutes and 30.69 minutes.

Example 2

Suppose we want to find the confidence interval for the average diameter of the regular M&M chocolate. For this purpose we randomly select the 20 M&M and find the following descriptive statistics:

Sample mean = Xbar = 1.018 cm

Sample standard deviation = S = 0.025 cm

Sample size = n = 20

df = n – 1 = 20 – 1 = 19

We assume 95% confidence level.

Critical t value = 2.0930

(By using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 1.018 ± 2.0930*0.025 /sqrt(20)

Confidence interval = 1.018 ± 2.0930*0.00559017

Confidence interval = 1.018 ± 0.0117

Lower limit = 1.018 - 0.0117 = 1.0063

Upper limit = 1.018 + 0.0117= 1.0297

WE are 95% confident that the average population diameter of the regular M&M chocolate will lies within 1.0063 cm and 1.0297 cm.