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.)

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

location: point 0

step-by-step explanation:

well, -8 to 10 on a number line is 18 spaces.

1/9 of 18 is 2.

so, 4/9 of 18 must be 8.

8 spaces from -8 (a) is 0.

so, 4/9 of the way from -8 to 10 is 0!