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? (assume the population data is normally distributed)

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]RequiredDetermine the test statisticLet 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 6For 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 deviationThis 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