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 15 12 22 27 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.)

95% confidence interval for the mean number of months is between a lower limit of 6.67 months and an upper limit of 25.73 months.

Step-by-step explanation:

Confidence interval is given as mean +/- margin of error (E)

Data: 5, 15, 12, 22, 27

mean = (5+15+12+22+27)/5 = 81/5 = 16.2 months

sd = sqrt[((5-16.2)^2 + (15-16.2)^2 + (12-16.2)^2 + (22-16.2)^2 + (27-16.2)^2) ÷ 5] = sqrt(58.96) = 7.68 months

n = 5

degree of freedom = n-1 = 5-1 = 4

confidence level (C) = 95% = 0.95

significance level = 1 - C = 1 - 0.95 = 0.05 = 5%

critical value (t) corresponding to 4 degrees of freedom and 5% significance level is 2.776

E = t×sd/√n = 2.776×7.68/√5 = 9.53 months

Lower limit of mean = mean - E = 16.2 - 9.53 = 6.67 months

Upper limit of mean = mean + E = 16.2 + 9.53 = 25.73 months

95% confidence interval is (6.67, 25.73)

answer: r there any options?

step-by-step explanation: