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: Exam Image Exam Image 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:

$t = 31.29$

Step-by-step explanation:

Given

$\begin{array}{ccccccc}{Stick} & {25.8} & {26.9} & {26.2} & {25.3} & {26.7}& {26.1} \ \\ {Liquid} & {16.9} & {17.4} & {16.8} & {16.2} & {17.3}& {16.8} \ \end{array}$

Required

Determine the test statistic

Let the dataset of stick be A and Liquid be B.

We start by calculating the mean of each dataset;

$\bar x =\frac{\sum x}{n}$

n, in both datasets in 6

For A

$\bar x_A =\frac{25.8+26.9+26.2+25.3+26.7+26.1}{6}$

$\bar x_A =\frac{157}{6}$

$\bar x_A =26.17$

For B

$\bar x_B =\frac{16.9+17.4+16.8+16.2+17.3+16.8}{6}$

$\bar x_B =\frac{101.4}{6}$

$\bar x_B =16.9$

Next, calculate the sample standard deviation

This is calculated using:

$s = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}$

For A

$s_A = \sqrt{\frac{\sum(x - \bar x_A)^2}{n-1}}$

$s_A = \sqrt{\frac{(25.8-26.17)^2+(26.9-26.17)^2+(26.2-26.17)^2+(25.3-26.17)^2+(26.7-26.17)^2+(26.1-26.17)^2}{6-1}}$

$s_A = \sqrt{\frac{1.7134}{5}}$

$s_A = \sqrt{0.34268}$

$s_A = 0.5854$  

For B

$s_B = \sqrt{\frac{\sum(x - \bar x_B)^2}{n-1}}$

$s_B = \sqrt{\frac{(16.9 - 16.9)^2+(17.4- 16.9)^2+(16.8- 16.9)^2+(16.2- 16.9)^2+(17.3- 16.9)^2+(16.8- 16.9)^2}{6-1}}$

$s_B = \sqrt{\frac{0.92}{5}}$

$s_B = \sqrt{0.184}$

$s_B = 0.4290$

Calculate the pooled variance

$S_p^2 = \frac{(n_A - 1)*s_A^2 + (n_B - 1)*s_B^2}{(n_A+n_B-2)}$

$S_p^2 = \frac{(6 - 1)*0.5854^2 + (6 - 1)*0.4290^2}{(6+6-2)}$

$S_p^2 = \frac{2.6336708}{10}$

$S_p^2 = 0.2634$

Lastly, calculate the test statistic using:

$t = \frac{(\bar x_A - \bar x_B) - (\mu_A - \mu_B)}{\sqrt{S_p^2/n_A +S_p^2/n_B}}$

We set

$\mu_A = \mu_B$

So, we have:

$t = \frac{(\bar x_A - \bar x_B) - (\mu_A - \mu_A)}{\sqrt{S_p^2/n_A +S_p^2/n_B}}$

$t = \frac{(\bar x_A - \bar x_B) }{\sqrt{S_p^2/n_A +S_p^2/n_B}}$

The equation becomes

$t = \frac{(26.17 - 16.9) }{\sqrt{0.2634/6 +0.2634/6}}$

$t = \frac{9.27}{\sqrt{0.0878}}$

$t = \frac{9.27}{0.2963}$

$t = 31.29$

The test statistic is 31.29