Question

5) A university wants to know more about the knowledge of students regarding international events. They are concerned that their students are uninformed in regards to news from other countries. A standardized test is used to assess student’s knowledge of world events (national reported mean=65, standard deviation=5). A sample of 30 students are tested (sample mean=58, standard deviation =3.2). Compute a 99 percent confidence interval based on this sample's data. How do these students compare to the national sample?

Answer 1

Solution:

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± Z*σ/sqrt(n)

From given data, we have

Xbar = 58

σ = 5

n = 30

Confidence level = 99%

Critical Z value = 2.5758

(by using z-table)

Confidence interval = Xbar ± Z*σ/sqrt(n)

Confidence interval = 58 ± 2.5758*5/sqrt(30)

Confidence interval = 58 ± 2.3514

Lower limit = 58 - 2.3514 = 55.65

Upper limit = 58 + 2.3514 =60.35

Confidence interval = (55.65, 60.35)

From this confidence interval, it is observed that the students score is less than the national average.