This hypothesis test 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.
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 null hypothesis would be that there is no significant evidence to conclude that the number of hospital admissions from motor vehicle crashes on Friday the 6th is different from the number of hospital admissions from motor vehicle crashes on Friday the 13th. That is, when the 13th day of a month falls on a Friday, the numbers of hospital admissions from motor vehicle crashes are not affected.
The alternative hypothesis is that there is significant evidence to conclude that the number of hospital admissions from motor vehicle crashes on Friday the 6th is different from the number of hospital admissions from motor vehicle crashes on Friday the 13th. That is, when the 13th day of a month falls on a Friday, the numbers of hospital admissions from motor vehicle crashes are affected.
If the population mean of number of hospital admissions from motor vehicle crashes on Friday the 13th = μ₁
And the population mean of number of hospital admissions from motor vehicle crashes on Friday the 6th = μ₂
If the difference in population mean of number of hospital admissions from motor vehicle crashes on Friday the 13th and number of hospital admissions from motor vehicle crashes on Friday the 6th is μ
μ = μ₁ - μ₂
Mathematically,
The null hypothesis is that
H₀: μ = 0
Or
H₀: μ₁ = μ₂
The alternative hypothesis is given as
Hₐ: μ ≠ 0
Or
Hₐ: μ₁ ≠ μ₂
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
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
To do this test, we will use the t-distribution because no information on the population standard deviation is known
So, we compute the t-test statistic
t = (x - μ)/σₓ
x = sample mean of difference = 6.80
μ = 0 (should be 0 if there's no difference between the two sets of data)
σₓ = standard error = [σ√n]
where n = Sample size = 5
σ = 4.12
σₓ = [4.12/√5] = 1.84
t = (6.80 - 0) ÷ 1.84
t = 3.70
checking the tables for the p-value of this t-statistic
Degree of freedom = df = n - 1 = 5 - 1 = 4
Significance level = 0.05
The hypothesis test uses a two-tailed condition because we're testing in two directions.
p-value (for t = 3.70, at 0.05 significance level, df = 4, with a two tailed condition) = 0.020835
The interpretation of p-values is that
When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.
So, for this question, significance level = 0.05
p-value = 0.020835
0.020835 < 0.05
Hence,
p-value < significance level
This means that reject the null hypothesis, accept the alternative hypothesis & say that there is enough evidence to conclude that number of hospital admissions from motor vehicle crashes on Friday the 6th is different from the number of hospital admissions from motor vehicle crashes on Friday the 13th. That is, when the 13th day of a month falls on a Friday, the numbers of hospital admissions from motor vehicle crashes are affected.
Hope this Helps!!!
