Answer:
Step-by-step explanation:
We would determine the mean and standard deviation for the stick and liquid fats.
Mean = sum of items/number of I teams
For the stick margarine,
n = 6
Mean = (25.5 + 26.7 + 26.5 + 26.6 + 26.3 + 26.4)/6 = 26.3
Standard deviation, s1 = √(summation(x - mean)^2/n
Summation(x - mean)^2 = (25.5 - 26.3)^2 + (26.7 - 26.3)^2 + (26.5 - 26.3)^2 + (26.6 - 26.3)^2 + (26.3 - 26.3)^2 + (26.4 - 26.3)^2 = 0.94
s1 = √0.94/6 = 0.41
For liquid margarine,
n = 6
Mean = (16.5 + 17.1 + 17.5 + 17.3 + 17.2 + 16.7)/6 = 17.1
Summation(x - mean)^2 = (16.5 - 17.1)^2 + (17.1 - 17.1)^2 + (17.5 - 17.1)^2 + (17.3 - 17.1)^2 + (17.2 - 17.1)^2 + (16.7 - 17.1)^2 = 0.73
s2 = √0.73/6 = 0.4
Since we know the sample standard deviation and the sample sizes are small, we would determine the t test statistic by applying the formula,
(x1 - x2)/√(s1²/n1 + s2²/n2)
x1 and x2 are the sample means
Therefore,
t = (26.3 - 17.1)/√(0.41²/6 + 0.4²/6)
t = 39.34
