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.

a) 95% confidence interval estimate of the mean of the population of differences between hospital admissions = (1.69, 11.91)

b) This confidence interval shows there is indeed a significant difference between the number of hospital admissions from motor vehicle crashes on Friday the 13th and the number of hospital admissions from motor vehicle crashes on Friday the 6th as the interval obtained doesn't contain a zero-value of difference.

Hence, 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 is not true.

Step-by-step explanation:

The missing data from the question

The numbers of hospital admissions from motor vehicle crashes

Friday the 6th || 10 | 8 | 4 | 4 | 2

Friday the 13th | 12 | 10 | 12 | 14 | 14

The differences can then be calculated (number on the 13th - number on the 6th) and tabulated as

Friday the 6th || 10 | 8 | 4 | 4 | 2

Friday the 13th | 12 | 10 | 12 | 14 | 14

Differences ||| 2 | 2 | 8 | 10 | 12

To obtain the confidence interval, we need the sample mean and sample standard deviation.

Mean = (Σx)/N

= (2+2+8+10+12)/5 = 6.80

Standard deviation = σ = √[Σ(x - xbar)²/N]

Σ(x - xbar)² = (2-6.8)² + (2-6.8)² + (8-6.8)² + (10-6.8)² + (12-6.8)² = 84.8

σ = √[Σ(x - xbar)²/N] = √(84.8/5) = 4.12

Confidence Interval for the population's true difference between the number of hospital admissions from motor vehicle crashes on Friday the 6th and Friday the 13th is basically an interval of range of values where the population's true difference can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample true difference) ± (Margin of error)

Sample Mean = 6.8

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the sample true difference)

Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 5 - 1 = 4.

Significance level for 95% confidence interval

(100% - 95%)/2 = 2.5% = 0.025

t (0.025, 4) = 2.776 (from the t-tables)

Standard error of the mean = σₓ = (σ/√n)

σ = standard deviation of the sample = 4.12

n = sample size = 5

σₓ = (4.12/√5) = 1.84

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 6.8 ± (2.776 × 1.84)

CI = 6.8 ± 5.10784

95% CI = (1.69216, 11.90784)

95% Confidence interval = (1.69, 11.91)

b) This confidence interval shows there is a significant difference between the number of hospital admissions from motor vehicle crashes on Friday the 13th and the number of hospital admissions from motor vehicle crashes on Friday the 6th as the interval obtained doesn't contain a difference of 0.

Hope this Helps!!!

i am thinking c. sorry if im wrong .

step-by-step explanation: