Question

Data on the numbers of hospital admissions resulting from motor vehicle crashes are given below for Fridays on the 6th of a month and Fridays on the following 13th of the same month. Assume that the paired sample data is a simple random sample and that the differences have a distribution that is approximately normal. Construct a​ 95% confidence interval estimate of the mean of the population of differences between hospital admissions. Use the confidence interval to test the claim that when the 13th day of a month falls on a​ Friday, the numbers of hospital admissions from motor vehicle crashes are not affected.

Friday the 6th	4	12	5	11	7
Friday the 13th	12	15	14	13	15

In this​ example, mu Subscript dμd is the mean value of the differences d for the population of all pairs of​ data, where each individual difference d is defined as the number of hospital admissions on Friday the 6th minus the number of hospital admissions on Friday the 13th. Find the​ 95% confidence interval.

____ < μd <_____

​(Round to two decimal places as​ needed.)

Answer 1

Solution:

Confidence interval for difference between mean of paired samples is given as below:

Confidence interval = Dbar ± t*S_D/sqrt(n)

From given data, we have

Dbar = -6

S_D = 3.2404

n = 5

df = n – 1 = 4

Confidence level = 95%

Critical t value = 2.7764

(by using t-table)

Confidence interval = Dbar ± t*S_D/sqrt(n)

Confidence interval = -6 ± 2.7764*3.2404/sqrt(5)

Confidence interval = -6 ± 4.0235

Lower limit = -6 - 4.0235 = -10.0235

Upper limit = -6 + 4.0235 = -1.9765

Confidence interval = (-10.0235, -1.9765)

-10.02 < µ_d < -1.98