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:[26.2,25.6,25.5,26.1,26.5,26.7] Liquid:[16.3,16.4,16.3,16.7,16.6,17.7] 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?

Answer 1

Answer:

t= 32.327

Would be a significant difference in the average amount of saturated fat in solid and liquid fats.

Step-by-step explanation:

1) Data given and notation

Stick:[26.2,25.6,25.5,26.1,26.5,26.7]

Liquid:[16.3,16.4,16.3,16.7,16.6,17.7]

$\bar X_{stick}$ represent the mean for the sample Stick

$\bar X_{Liquid}$ represent the mean for the sample Liquid

$s_{stick}$ represent the sample standard deviation for the sample Stick

$s_{Liquid}$ represent the sample standard deviation for the sample Liquid

$n_{stick}=6$ sample size for the group Stick

$n_{Liquid}=6$ sample size for the group Liquid

t would represent the statistic (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:

Null Hypothesis :$\mu_{stick}=\mu_{Liquid}$

Alternative Hypothesis :$\mu_{stick} \neq \mu_{Liquid}$

If we analyze the size for the samples both are < 30 so for this case we can apply a t test to compare means, and the statistic formula is:

$t=\frac{\bar X_{stick}-\bar X_{Liquid}}{\sqrt{\frac{s^2_{stick}}{n_{stick}}+\frac{s^2_{Liquid}}{n_{Liquid}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)  

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)  

3) Calculate the statistic

First we need to calculate the mean and deviation for each sample, after apply the formulas (2) and (3) we got the following results:

$\bar X_{stick}=26.10$ $s_{stick}=0.477$

$\bar X_{Liquid}=16.67$ $s_{Liquid}=0.532$

And with this we can replace in formula (1) like this:

$t=\frac{26.10-16.67}{\sqrt{\frac{0.477^2}{6}+\frac{0.532^2}{6}}}}=32.327$  

4) Statistical decision

For this case we don't have a significance level provided $\alpha$, but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:

$df=n_{stick}+n_{liquid}-2=6+6-2=10$

Since is a bilateral test the p value would be:

$p_v =2*P(t_{(10)}>32.327)=1.8x10^{-11}$

So the p value is a very low value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and a would be a significant difference in the average amount of saturated fat in solid and liquid fats.