Oil content of fried sweet potato chips. The characteristics of sweet potato chips fried at different temperatures were investigated in the Journal of Food Engineering (September 2013). A sample of 6 sweet potato slices was fried at 130° using a vacuum fryer. One characteristic of interest to the researchers was internal oil content (measured in millions of grams). The results were: $\overline{x}=178$and $s=11$. The researchers are interested in estimating the variance of the interval oil content measurements for sweet potato chips.

a. Identify the target parameter, in symbols and words.

b. Compute a 95% confidence interval for ${\sigma }^2$.

c. What does it mean to say that the target parameter lies within the interval with “95% confidence”?

d. What assumption about the data must be satisfied in order for the confidence interval to be valid?

e. To obtain a practical interpretation of the interval, part b, explain why a confidence interval for the standard deviation, $\sigma$, is desired.

f. Use the results, part b, to compute a 95% confidence interval for$\sigma$ . Give a practical interpretation of the interval.

A sample of 6 sweet potato slices was fried at 130° using a vacuum fryer. So, the sample size n=6. Also, sample mean and standard deviation is given as $\overline{x}=178$and$s=11$

In the study, the parameter is that the population variation of the oil content measurements for sweet potato chips, which is denoted as${\sigma }^2$

Since, the $100\left(1-\alpha \right)\%$confidence interval is given as,

$\left(\frac{\left(n-1\right)s^2}{{\chi }_{0.05}^2}⩽{\sigma }^2⩽\frac{\left(n-1\right)s^2}{{\chi }_{0.95}^2}\right)$

Let the confidence level be,

$\begin{array}{rcl}1-\alpha & =& 0.95\\ \alpha & =& 1-0.95\\ & =& 0.05\\ & & \end{array}$

Therefore,

$\frac{\alpha }{2}=0.025$

Also, the degrees of freedom is $\left(n-1\right)=5$

Now, from table p-values are,

$$\begin{gathered} {\chi }_{0.025}^2\left(5\right)=12.833 \\ {\chi }_{0.975}^2\left(5\right)=0.831 \end{gathered}$$

Therefore,

$\begin{array}{rcl}\left(\frac{\left(n-1\right)s^2}{{\chi }_{0.025}^2}⩽{\sigma }^2⩽\frac{\left(n-1\right)s^2}{{\chi }_{0.975}^2}\right)& =& \left(\frac{\left(6-1\right)121}{12.833}⩽{\sigma }^2⩽\frac{\left(6-1\right)121}{0.831}\right)\\ & =& \left(\frac{\left(5\right)121}{12.83}⩽{\sigma }^2⩽\frac{\left(5\right)121}{0.831}\right)\\ & =& \left(47.15⩽{\sigma }^2⩽728.04\right)\\ & & \end{array}$

Hence, the 95% confidence interval for${\sigma }^2$ is (47.15, 728.04).

The 95% of all similarly constructed confidence intervals will contain the true value of population variancein repeated sampling.

The general conditions required for a valid confidence interval for are given below:

• The sample is randomly selected from the population.
• The population follows normal distribution

The assumptions required for the interval to be valid is that the population of oil content measurements for sweet potato chips follows normal distribution and the sample is randomly selected from the population.

e.

In the study, the measurement of variance is square millions of grams, which is not easy to connect with the given data. Moreover, the measurement of standard deviation is millions of grams, which are easy to connect with the given data. Hence, the confidence interval for the standard deviation,, $\sigma$is desired.

From part b, the 95% confidence interval is (47,728). Take the square root on the lower and upper limit. That is,

$\left(\sqrt{47.15},\sqrt{728.04}\right)=\left(6.87,26.98\right)$

There is 95% confident that the population standard deviation of oil content measurements for sweet potato chips lies between the interval 6.9 and 27.