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	2	3	8	8	10
Friday the 13th	15	14	14	13	12

In this​ example,

mu Subscript d μ d 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.

Answer 1

Solution:

The required confidence interval is given as below:

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

From given data, we have

Dbar = -7.4

S_d = 4.5056

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 = -7.4 ± 2.7764*4.5056/sqrt(5)

Confidence interval = -7.4± 2.7764*2.014965576

Confidence interval = -7.4 ± 5.5944

Lower limit = -7.4 - 5.5944 = -12.9944

Upper limit = -7.4 + 5.5944= -1.8056

Confidence interval = (-12.9944, -1.8056)

When the 13th day of a month falls on a​ Friday, the numbers of hospital admissions from motor vehicle crashes are affected, because above confidence interval does not zero.