Since we do not know the population standard deviation, we will perform a one-sample t-test.
Calculating the test statistic will require calculating the sample mean, which can be done as follows:
Sample mean (x bar over it) = (90.4 + 116.4 + 85.9 + 90.4)/4 ~ 95.775.
Sample standard deviation (s) = √[(Σ(xi - x̄)^2) / (n - 1)]
Sample standard deviation (s) = √[((90.4 - 95.775)^2 + (116.4 - 95.775)^2 + (85.9 - 95.775)^2 + (90.4 - 95.775)^2) / (4 - 1)]
Sample standard deviation (s) ≈ √[(27.4194 + 441.1349 + 99.0156 + 27.4194) / 3]
Sample standard deviation (s) ≈ √[595.9893 / 3]
Sample standard deviation (s) ≈ √198.6631
Sample standard deviation (s) ≈ 14.100 (rounded to three decimal places)
Step 2: Calculate the test statistic (t).
The formula for the one-sample t-test is:
t = (x̄ - μ) / (s / √n)
where:
x̄ = Sample mean
μ = Population mean under the null hypothesis (51.6 in this case)
s = Sample standard deviation
n = Sample size (number of data points)
t = (95.775 - 51.6) / (14.100 / √4)
t ≈ 44.175 / (14.100 / 2)
t ≈ 44.175 / 7.050
t ≈ 6.257 (rounded to three decimal places)
Step 3: Calculate the degrees of freedom (df).
Since we have a sample size of 4, the degrees of freedom is given by (n - 1) = (4 - 1) = 3.
Step 4: Calculate the p-value.
Using a t-table or statistical software, we find that the p-value for a t-statistic of 6.257 with 3 degrees of freedom is approximately 0.0033 (rounded to four decimal places).
Therefore, the test statistic for this sample is 6.257, and the p-value is 0.0033.
