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Which Method? Refer to Exercise 7 鈥淩equirements鈥 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 2distribution? If not, what other method could be used to find a 95% confidence interval estimate of.

Short Answer

Expert verified

The 95% confidence interval to estimate cannot be computed using the 2distribution 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 as this method does not require the sample to come from a normally distributed population.

Step by step solution

01

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.

02

Appropriate method

A strict requirement to compute the confidence interval estimate of using the2distribution 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 cannot be computed using the 2distribution.

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 will be expressed as shown below:

P2100<<P1-2100=P0.052100<<P1-0.052100

Thus, the limits are obtained as follows:

P2.5<<P97.5

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