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 10    7 11 9 9
Friday the 13th    10    12 11    15    15
In this example, μd 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.)
Based on the confidence interval, can one reject 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?
A. No, because the confidence No interval includes zero.
B. No, because the confidence interval does not include zero.
C. Yes, because the confidence interval does not include zero.
D. Yes, because the confidence interval includes zero

Answer 1

Given :

Friday the 6th	10	7	11	9	9
Friday the 13th	10	12	11	15	15

μd is the mean value of the differences d for the population of all pairs of data

d = Friday the 6th - Friday the 13th

claim : the 13th day of a month falls on a Friday, the numbers of hospital admissions from motor vehicle crashes are not affected.

d = Friday the 6th - Friday the 13th

So ,

Now the confidence interval formula for the mean of the population of differences is ,

We have to find the 95% confidence interval ,

c = 0.95

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

df = 4

Using Excel function ,   =TINV( , df )

=TINV( 0.05 , 4 )

=2.776

Now plug the values in the formula ,

Rounding to two decimals would be ,

Therefore the 95% confidence interval estimate of the mean of the population of differences between hospital admissions is ( -7.29 , 0.49 )

As zero (0) is included in the confidence interval , we fail to reject the null hypothesis (Ho).

That is we can not reject 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 .

So the correct answer choice is , A. No, because the confidence interval includes zero.

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Note : In answer choice A. No, because the confidence No interval includes zero.

i think the underlined No is typo.