Question

Which Method? Refer to Exercise 7 “Requirements” and assume that sample of 12 voltage levels appears to be from a population with a distribution that is substantially far from being normal. Should a 95% confidence interval estimate ofbe constructed using the ${\chi }^2$distribution? If not, what other method could be used to find a 95% confidence interval estimate of$\sigma$.

Short answer

The 95% confidence interval to estimate cannot be computed using the ${\chi }^2$distribution because the sample of voltage levels does not come from a population that is normally distributed.

The bootstrap method can be adopted to find a 95% confidence interval to estimate $\sigma$as this method does not require the sample to come from a normally distributed population.

Step-by-step solution

Given information

It is given that a sample of 12 voltage levels of smartphone batteries comes from a population that is normally distributed. A 95% confidence interval is to be constructed to estimate the standard deviation of voltage levels.

Appropriate method

A strict requirement to compute the confidence interval estimate of $\sigma$using the${\chi }^2$distribution is that the sample should be selected from a normally distributed population (even if the sample size is large).

As the sample of voltage levels does not come from a population that is normally distributed, the 95% confidence interval to estimate $\sigma$cannot be computed using the ${\chi }^2$distribution.

An alternate method that can be used to compute the confidence interval is discussed below:

• Obtain a set of 1000 or more bootstrap samples of size n=12 from the given sample.
• A bootstrap sample is a random sample obtained with the replacement of values from the given sample.
• Compute the sample standard deviation for each of the bootstrap samples.

• Arrange the set of all sample standard deviations in ascending order.

• Construct the confidence interval by computing the suitable percentile values. Here, the 95% confidence interval to estimate $\sigma$will be expressed as shown below:

$P_{\frac{\alpha }{2}\times 100}<\sigma <P_{1-\frac{\alpha }{2}\times 100}=P_{\frac{0.05}{2}\times 100}<\sigma <P_{1-\frac{0.05}{2}\times 100}$

Thus, the limits are obtained as follows:

$P_{2.5}<\sigma <P_{97.5}$