Question

Solid fats are more likely to raise blood cholesterol levels than

liquid fats. Suppose a nutritionist analyzed the percentage of saturated fat for a sample of 6 brands of stick margarine (solid fat) and for a sample of 6 brands of liquid margarine and obtained the following results:
Stick= [25.5, 26.7, 26.5, 26.6, 26.3, 26.4]
Liquid= [16.5, 17.1, 17.5, 17.3, 17.2, 16.7]
We want to determine if there a significant difference in the average amount of saturated fat in solid and liquid fats. What is the test statistic? (assume the population data is normally distributed)
a. t = 25.263
b. z = 39.604
c. t = 39.604
d. t = 39.104
e. z = 39.104

Answer 1

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

Answer 2

c

Explanation:

mean of stick= 26.33

mean of liquid= 17.05

standard deviation of stick= 0.394

standard deviation of liquid= 0.345

Here we are comparing two samples so we will use t-test

t-statistic= (mean1-mean2)/√(SD1²/(N1-1) + SD2²/(N2-1))

= 39.6