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Researchers collected data on the numbers of hospital admissions resulting from motor vehicle crashes, and results are given below for Fridays on the 6 th of a month and Fridays on the following 13 th of the same month (based on data from "Is Friday the 13 th Bad for Your Health?" by Scanlon et al., British Medical Journal, Vol. 307, as listed in the Data and Story Line online resource of data sets). Construct a \(95 \%\) confidence interval estimate of the mean of the population of differences between hospital admissions on days that are Friday the 6 th of a month and days that are Friday the 13 th of a month. Use the confidence interval to test the claim that when the 13 th day of a month falls on a Friday, the numbers of hospital admissions from motor vehicle crashes are not affected. $$ \begin{array}{|l|r|r|r|r|r|r|} \hline \text { Friday the 6th } & 9 & 6 & 11 & 11 & 3 & 5 \\ \hline \text { Friday the 13th } & 13 & 12 & 14 & 10 & 4 & 12 \\ \hline \end{array} $$

Step by step solution

01

- Calculate the Differences

First, find the differences between hospital admissions on Friday the 6th and Friday the 13th for each pair of dates. Subtract each Friday the 6th value from the corresponding Friday the 13th value: \[ 13 - 9 = 4 \]\[ 12 - 6 = 6 \]\[ 14 - 11 = 3 \]\[ 10 - 11 = -1 \]\[ 4 - 3 = 1 \]\[ 12 - 5 = 7 \]

02

- Calculate the Mean and Standard Deviation of Differences

Calculate the mean (\( \bar{d} \)) and the standard deviation (\( s_d \)) of the differences: \[ \bar{d} = \frac{4 + 6 + 3 - 1 + 1 + 7}{6} = \frac{20}{6} = 3.33 \] To calculate the standard deviation (\( s_d \)): \[ s_d = \sqrt{\frac{(4-3.33)^2 + (6-3.33)^2 + (3-3.33)^2 + (-1-3.33)^2 + (1-3.33)^2 + (7-3.33)^2}{6-1}} = \sqrt{\frac{25.34}{5}} = 2.25 \]

03

- Find the t-Score for the Confidence Interval

Using a t-distribution table and a 95% confidence level with \( n-1 = 5 \) degrees of freedom, find the t-score (\( t_{0.025, 5} \)). The t-score is approximately 2.571.

04

- Calculate the Margin of Error

Calculate the margin of error (ME) using the formula: \[ ME = t_{\alpha/2} \times \frac{s_d}{\sqrt{n}} = 2.571 \times \frac{2.25}{\sqrt{6}} = 2.36 \]

05

- Construct the Confidence Interval

Construct the confidence interval using the formula: \[ \bar{d} \pm ME = 3.33 \pm 2.36 \] So, the 95% confidence interval is: \( [0.97, 5.69] \)

06

- Interpret the Confidence Interval

Since the 95% confidence interval \( [0.97, 5.69] \) does not contain 0, it suggests that there is a significant difference in hospital admissions on Fridays the 6th and Fridays the 13th. Thus, it supports the claim that the number of hospital admissions from motor vehicle crashes is affected when the 13th day of a month falls on a Friday.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean Difference

The mean difference is a measure that helps to understand the average change between two sets of data. In this exercise, we are comparing hospital admissions on Friday the 6th and Friday the 13th. We first calculate the difference between each pair of corresponding values. Here, the differences are: 4, 6, 3, -1, 1, and 7. The mean of these differences (denoted as \( \bar{d} \)) is then calculated to understand the average difference. Mathematically, the mean difference is the sum of all individual differences divided by the number of observations: \[ \bar{d} = \frac{4 + 6 + 3 - 1 + 1 + 7}{6} = \frac{20}{6} = 3.33 \] This result means that, on average, there are 3.33 more hospital admissions on Friday the 13th compared to Friday the 6th.

Standard Deviation

The standard deviation (denoted as \( s_d \)) measures the amount of variation or dispersion in a set of values. A low standard deviation means that the values tend to be close to the mean, while a high standard deviation indicates that they are spread out over a wider range. To calculate the standard deviation of the differences, we follow these steps:

• Find the deviation of each difference from the mean difference.
• Square each deviation.
• Sum all squared deviations.
• Divide this sum by the number of observations minus one (this is called the variance).
• Take the square root of the variance to get the standard deviation.

For our data, we get: \[ s_d = \frac{1}{\text{n}-1}\boldsymbol{\text{Var}(x) + \text{Var}(y)} +1] = \frac{25.34}{5} = 2.25 \] This indicates that the individual differences vary quite a bit from the mean difference.

t-Distribution

The t-distribution is a type of probability distribution that is used when the sample size is small and/or the population standard deviation is unknown. It is similar to the normal distribution but has heavier tails, which means it is more prone to producing values that fall far from its mean. In this exercise, we use the t-distribution to find the t-score that corresponds to our desired confidence level (95%) and degrees of freedom (n-1). Using a t-table, we find the t-score (denoted as \( t_{\text{0.025, 5}} \)) to be approximately 2.571. This t-score is critical because it will be used in calculating the margin of error in our confidence interval.

Margin of Error

The margin of error (ME) represents the range within which the true population parameter is expected to lie, based on our sample data. It is calculated using the t-score, standard deviation of the differences, and the number of observations. The formula for the margin of error is: \[ ME = t_{\frac{\text{α}}{2} } \times \frac{s_d}{\text{number of observations}} \] For our data, this translates to: \[ ME = 2.571 \times \frac{2.25}{\textOnumber of values} = 2.36 \] This margin of error tells us how much error we can expect in our estimate of the mean difference.

Statistical Significance

Statistical significance helps us determine whether the observed differences or relationships in our data occurred by chance or are indeed meaningful. In this exercise, we use the 95% confidence interval to test the claim. Our calculated confidence interval is \([0.97, 5.69]\), which does not contain 0. This suggests a significant difference in hospital admissions on Fridays the 6th and Fridays the 13th. If the interval had contained 0, we could not have rejected the null hypothesis that there is no significant difference. This result supports the claim that the number of hospital admissions from motor vehicle crashes is affected when the 13th day of a month falls on a Friday.