Ask Question
4 November, 09:58

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.)

+4
Answers (1)
  1. 4 November, 10:28
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Five students visiting the student health center for a free dental examination during national dental hygiene month were asked how many ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers