Question

Oil content of fried sweet potato chips. Refer to theJournal of Food Engineering (September 2013) study of the characteristics of fried sweet potato chips, Exercise 7.90 (p. 431). Recall that a sample of 6 sweet potato slices fried at 130° using a vacuum fryer yielded the following statistics on internal oil content (measured in gigagrams [Gg]): x1 = .178 Gg and s1 = .011 Gg. A second sample of 6 sweet potato slices was obtained, but these were subjected to a two-stage frying process (again, at 130°) in an attempt to
improve texture and appearance. Summary statistics on internal oil content for this second sample follows: x2 = .140 Gg and s2 = .002 Gg. Using a t-test, the researchers want to compare the mean internal oil content of sweet potato chips fried with the two methods. Do you recommend the researchers carry out this analysis? Explain.

Short answer

The answer can be reduced from the following steps.

Step-by-step solution

Given information

$$\begin{gathered} n_1=6 \\ {n}_2=6 \\ {\overline{x}}_1=.178Gg \\ {\overline{x}}_2=.140Gg \\ {S}_1=.011Gg \\ {S}_2=.002Gg \end{gathered}$$

Explaining the t-test

T-test: The t-test is a study that provides empirical used to assess the relationship between two groups' means and assess whether there is a significant difference between them.

When data sets have unknown variances and a normal distribution, such as the data set obtained from tossing a coin 100 times, t-tests are used.

Application of t-test

Due to the two sample sizes, two variances, and two standard deviations, the T-test is the most recommended. Therefore, this test evaluates the outcomes of two groups. Other tests, however, only evaluate one group.

The t-test is as follows.

$t=\frac{Difference between mean}{\frac{Variance}{Sample size}}$

$t=\frac{\left(\overline{x}1-\overline{x}2\right)}{\sqrt{\frac{S^{2}1}{n1}+\frac{S^2}{n2}}}$

$t=\frac{0.038}{\sqrt{\frac{0.000121}{6}+\frac{0.000004}{6}}}$

$$\begin{gathered} t=\frac{0.038}{\sqrt{\frac{0.00121}{6}+\frac{0.000004}{6}}} \\ t=\frac{0.038}{\sqrt{\frac{0.00726+0.0000024}{36}}} \end{gathered}$$

$$\begin{gathered} t=\frac{0.038}{\sqrt{\frac{0.007284}{36}}} \\ t=\frac{0.038}{\sqrt{0.00020233}} \\ t=\frac{0.038}{0.0142242} \\ t=2.67150349 \end{gathered}$$