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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

Sagot :

MrRoyalidn

Answer:

[tex]t = 31.29[/tex]

Step-by-step explanation:

Given

[tex]\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}[/tex]

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;

[tex]\bar x =\frac{\sum x}{n}[/tex]

n, in both datasets in 6

For A

[tex]\bar x_A =\frac{25.8+26.9+26.2+25.3+26.7+26.1}{6}[/tex]

[tex]\bar x_A =\frac{157}{6}[/tex]

[tex]\bar x_A =26.17[/tex]

For B

[tex]\bar x_B =\frac{16.9+17.4+16.8+16.2+17.3+16.8}{6}[/tex]

[tex]\bar x_B =\frac{101.4}{6}[/tex]

[tex]\bar x_B =16.9[/tex]

Next, calculate the sample standard deviation

This is calculated using:

[tex]s = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}[/tex]

For A

[tex]s_A = \sqrt{\frac{\sum(x - \bar x_A)^2}{n-1}}[/tex]

[tex]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}}[/tex]

[tex]s_A = \sqrt{\frac{1.7134}{5}}[/tex]

[tex]s_A = \sqrt{0.34268}[/tex]

[tex]s_A = 0.5854[/tex]  

For B

[tex]s_B = \sqrt{\frac{\sum(x - \bar x_B)^2}{n-1}}[/tex]

[tex]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}}[/tex]

[tex]s_B = \sqrt{\frac{0.92}{5}}[/tex]

[tex]s_B = \sqrt{0.184}[/tex]

[tex]s_B = 0.4290[/tex]

Calculate the pooled variance

[tex]S_p^2 = \frac{(n_A - 1)*s_A^2 + (n_B - 1)*s_B^2}{(n_A+n_B-2)}[/tex]

[tex]S_p^2 = \frac{(6 - 1)*0.5854^2 + (6 - 1)*0.4290^2}{(6+6-2)}[/tex]

[tex]S_p^2 = \frac{2.6336708}{10}[/tex]

[tex]S_p^2 = 0.2634[/tex]

Lastly, calculate the test statistic using:

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

We set

[tex]\mu_A = \mu_B[/tex]

So, we have:

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

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

The equation becomes

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

[tex]t = \frac{9.27}{\sqrt{0.0878}}[/tex]

[tex]t = \frac{9.27}{0.2963}[/tex]

[tex]t = 31.29[/tex]

The test statistic is 31.29

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