Question

Five students visiting the student health center for a free dental examination during national dental hygiene month were asked how many months had passed since their last visit to a dentist. their responses were as follows. 5 18 12 24 28 assuming that these five students can be considered a random sample of all students participating in the free checkup program, construct a 95% confidence interval for the mean number of months elapsed since the last visit to a dentist for the population of students participating in the program. (give the answer to two decimal places.)

Answer 1

To find for the value of the confidence interval, let us first calculate for the values of x and s, the mean and standard deviation respectively.

x = (5 + 18 + 12 + 24 + 28) / 5

x = 17.4 months

 

s = sqrt{[(5 – 17.4)^2 + (18 – 17.4)^2 + (12 – 17.4)^2 + (24 – 17.4)^2 + (28 – 17.4)^2]/(5-1)}

s = 9.21

 

The formula for the confidence interval is given as:

Confidence Interval = x ± t s / sqrt(n)

Where t can be taken from standard distribution tables at 95% level at degrees of freedom = n – 1 = 4, t = 2.132. Therefore:

Confidence Interval = 17.4 ± 2.132 * 9.21 / sqrt(5)

Confidence Interval = 17.4 ± 8.78

Confidence Interval = 8.62 months, 26.18 months