Question

Short answer

The 95% confidence interval is equal to (-6.49,-0.17).

There is sufficient evidence to 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.

Step-by-step solution

Given information

The data are given about the hospital admissions resulting from motor vehicle crashes on two days of the month, Friday the 6th and Friday the 13th.

Hypotheses

Null hypothesis: The mean of the population of differences in the number of hospital admissions on Friday the 6th and Friday the 13thequals 0.

\({H_{0\;}}:\;{\mu _d} = 0\)

Alternative hypothesis: The mean of the population of differences in the number of hospital admissions on Friday the 6th and Friday the 13th is not equal to 0.

\({H_1}\;:\;{\mu _d} \ne 0\)

Here,\({\mu _d}\;\)is the mean of the population of differences in the number of hospital admissions for the two days.

Table of differences

The following table shows thedifferences in the number of hospital admissions on Friday the 6th and Friday the 13th :

Friday the 6th	9	6	11	11	3	5
Friday the 13th	13	12	14	10	4	12
Differences (d)	-4	-6	-3	1	-1	-7

Mean of differences and standard deviation of differences

The value of the mean of the differences in the number of hospitaladmissions on Friday the 6th and Friday the 13^this computed below:

\(\begin{array}{c}\bar d = \frac{{\left( { - 4} \right) + \left( { - 6} \right) + ..... + \left( { - 7} \right)}}{6}\\ = - 3.33\end{array}\)

The value of the standard deviation of the differences in the number ofhospital admissions on Friday the 6th and Friday the 13th is computed below:

\(\begin{array}{c}{s_d} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({d_i} - \bar d)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {\left( { - 4} \right) - \left( { - 3.33} \right)} \right)}^2} + {{\left( {\left( { - 6} \right) - \left( { - 3.33} \right)} \right)}^2} + ...... + {{\left( {\left( { - 7} \right) - \left( { - 3.33} \right)} \right)}^2}}}{{6 - 1}}} \\ = 3.01\end{array}\)

Confidence Interval

The formula of the confidence interval is given below:

\(CI = \bar d - E < {\mu _d} < \bar d + E\;\)

Compute the margin of error

The level of significance is equal to 0.05.

The degrees of freedom are computed below:

\(\begin{array}{c}df = n - 1\\ = 6 - 1\\ = 5\end{array}\)

The value of the margin of error is equal tothe following:

\(\begin{array}{c}E = {t_{\frac{\alpha }{2},df}} \times \frac{{{s_d}}}{{\sqrt n }}\\ = {t_{0.025,5}} \times \frac{{3.01}}{{\sqrt 6 }}\\ = 2.5706 \times \frac{{3.01}}{{\sqrt 6 }}\\ = 3.15997\end{array}\)

Compute the confidence interval

Substituting the required values, the following value is obtained:

\(\begin{array}{c}\bar d - E < {\mu _d} < \bar d + E\;\\\left( { - 3.33 - 3.15997} \right) < {\mu _d} < \left( { - 3.33 + 3.15997} \right)\\ - 6.49 < {\mu _d} < - 0.17\end{array}\)

Thus, the 95% confidence interval equals (-6.49,-0.17).

Conclusion based on the confidence interval

Since the confidence interval does not contain the value 0, it can be said that the mean of the difference in the number of hospital admissions cannot be equal to 0.

Thus, there is sufficient evidence to 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.