Question

A sample of 58 CCA students show that the average age of CCA students is ¯xx¯ =22.4 years, with a standard deviation of ss = 4.4 years.


1. Compute the z-score for a student who is xx = 38.64 years old.
z-score=


2.Interpret the z-score from part 1.

3. Using the mean and standard deviation from above, construct a 94 % confidence interval for the average age of all CCA students.
Lower Bound =
Upper Bound =
4. If you decided to change the confidence level in #3 to 86 %. The margine of error will

Answer 1

Solution:

We are given

n = 58

Mean = Xbar = 22.4

SD = S = 4.4

Question 1

Z = (X – mean) /SD

We are given X = 38.64

Z = (38.64 - 22.4)/4.4

Z = 3.690909

Question 2

Interpretation of above z-score: The z score value of 3.69 is greater than 3 and therefore the age of the student is very high as compared to other students.

Question 3

Confidence interval for Population mean is given as below:

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

We are given

Confidence level = 94%

n = 58

Mean = Xbar = 22.4

SD = S = 4.4

df = n – 1 = 58 – 1 = 57

Critical t value = 1.9190

(by using t-table)

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

Confidence interval = 22.4 ± 1.9190*4.4/sqrt(58)

Confidence interval = 22.4 ± 1.9190*0.577748305

Confidence interval = 22.4 ± 1.1087

Lower Bound = 22.4 - 1.1087 = 21.29

Lower Bound = 21.29

Upper Bound = 22.4 + 1.1087 = 23.51

Upper Bound = 23.51

Question 4

Margin of error = t*S/sqrt(n)

We are given

Confidence level = 86%

S = 4.4

n = 58

df = n – 1 = 58 – 1 = 57

Critical t value = 1.4967

(by using t-table)

Margin of error = 1.4967*4.4/sqrt(58)

Margin of error = 1.4967*0.577748305

Margin of error = 0.8647