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. 6 19 10 24 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.) ( , )

In a study of academic procrastination, the authors of a paper reported that for a sample of 411 undergraduate students at a midsize public university preparing for a final exam in an introductory psychology course, the mean time spent studying for the exam was 7.34 hours and the standard deviation of study times was 3.90 hours. For purposes of this exercise, assume that it is reasonable to regard this sample as representative of students taking introductory psychology at this university.

(a) Construct a 95% confidence interval to estimate μ, the mean time spent studying for the final exam for students taking introductory psychology at this university. (Round your answers to three decimal places.)
(  ,  )

(b) The paper also gave the following sample statistics for the percentage of study time that occurred in the 24 hours prior to the exam.

n = 411      x = 43.98      s = 21.96

Construct a 90% confidence interval for the mean percentage of study time that occurs in the 24 hours prior to the exam. (Round your answers to three decimal places.)

Answer 1

Question 1

	Values ( X )	Σ ( Xi- X̅ )2
	6	125.44
	19	3.24
	10	51.84
	24	46.24
	27	96.04
Total	86	322.8

Part a)

Mean X̅ = Σ Xi / n
X̅ = 86 / 5 = 17.2
Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 322.8 / 5 -1 ) = 8.9833

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 5- 1 ) = 2.776
17.2 ± t(0.05/2, 5 -1) * 8.9833/√(5)
Lower Limit = 17.2 - t(0.05/2, 5 -1) 8.9833/√(5)
Lower Limit = 6.048
Upper Limit = 17.2 + t(0.05/2, 5 -1) 8.9833/√(5)
Upper Limit = 28.352
95% Confidence interval is ( 6.048 , 28.352 )

Part b)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.1 /2, 411- 1 ) = 1.649
43.98 ± t(0.1/2, 411 -1) * 21.96/√(411)
Lower Limit = 43.98 - t(0.1/2, 411 -1) 21.96/√(411)
Lower Limit = 42.194
Upper Limit = 43.98 + t(0.1/2, 411 -1) 21.96/√(411)
Upper Limit = 45.766
90% Confidence interval is ( 42.194 , 45.766 )