Do You Know How to Compute Confidence Intervals?

In this week's tutorial, we are going to be covering the topic of Confidence Intervals. See below a list of relevant sample problems, with step by step solutions.

Question 1: A type of lithium battery is being assessed for its length of life. The production manager selected a random sample of 10 batteries and recorded the following lifetimes in years: {3.25, 4.0, 3.1, 3.7, 3.5, 4.2, 4.75, 2.3, 5.5, 3.7}. Answer the following assuming that the population is normal.

a. What is the sample mean?

b. What is the sample standard deviation?

c. Explain how the sample mean is related to the population mean.

d. Assuming that you do not know the standard deviation of the population, construct a

90% confidence interval for $$mu$$ .

e. Assume that you do know that the standard deviation of the population is $$\sigma$$ = 0.7; construct a 90% confidence interval for $$\sigma$$ . (Show formula or calculator command)

f. Interpret the confidence interval in part e.

Solution: (a) The following table is provided

 Data 3.25 4 3.1 3.7 3.5 4.2 4.75 2.3 5.5 3.7 Mean 3.8 St. Dev 0.891

The sample mean is 3.8

(b) The sample standard deviation is 0.891.

(c)The sample mean is the point estimate for the population mean.

(d) The population standard deviation is unknown so we are going to use the t-statistics. The 90% confidence interval is given by

$CI=\left( \bar{X}-{{t}_{\alpha /2}}\times \frac{s}{\sqrt{n}},\,\,\bar{X}+{{t}_{\alpha /2}}\times \frac{s}{\sqrt{n}} \right)$

In this case we have $${{t}_{\alpha /2}}$$ is the two-tailed t-critical value, for $$\alpha =0.10$$ and $$n-1 = 9$$ degrees of freedom. Therefore, we obtain that

$CI=\left( {3.8}-1.833\times \frac{0.891}{\sqrt{10}},\,\,{3.8}+1.833\times \frac{0.891}{\sqrt{10}} \right)=\left( {3.2835},\,\,{4.3165} \right)$

The interpretation is that we are 90% confident that the actual population mean $$\mu$$ is contained by the interval $$\left( {3.2835},\,\text{ }{4.3165} \right)$$.

(d) The population standard deviation is available to, so the normal distribution can be used. Therefore, we get that the 90% confidence interval is given

$CI=\left( \bar{X}-{{z}_{\alpha /2}}\times \frac{\sigma}{\sqrt{n}},\,\,\bar{X}+{{z}_{\alpha /2}}\times \frac{\sigma}{\sqrt{n}} \right)$

where $${{z}_{\alpha /2}}$$ corresponds to the two-tailed z-critical value for $$\alpha =0.10$$. Therefore, we find that

$CI=\left( {3.8}-{1.6449}\times \frac{0.7}{\sqrt{10}},\,\,{3.8}+{1.6449}\times \frac{0.7}{\sqrt{10}} \right)=\left( {3.4359},\,\,{4.1641} \right)$

(e) The interpretation is that we are 90% confident that the actual population mean $$\mu$$ is contained by the interval (3.4359, 4.1641).

Question 2: A random sample of 56 fluorescent light bulbs has a mean life of 645 hours with a standard deviation of 31 hours. Construct a 95% confidence interval for the population mean.

Solution: The 95% confidence interval for the population mean is given by

$CI=\left( \bar{X}-{{z}_{\alpha /2}}\times \frac{s}{\sqrt{n}},\text{ }\bar{X}+{{z}_{\alpha /2}}\times \frac{s}{\sqrt{n}} \right)=\left( 645-1.96\times \frac{31}{\sqrt{56}},\text{ }645+1.96\times \frac{31}{\sqrt{56}} \right)$

$=\left( 636.8806,\text{ }653.1194 \right)$

Question 3: A simple random sample is to be drawn from a population of 1200. In order to have 90% confidence that the sampling error in estimating $$p$$ is no more than 0.03, what sample size will be necessary?

Solution: The 90% confidence interval is given by

$CI=\left( \hat{p}-{{z}_{\alpha /2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}},\text{ }\hat{p}+{{z}_{\alpha /2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \right)$

where $${{z}_{\alpha /2}}=1.645$$. Therefore, the margin of error is

$MOE=1.645\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

We want the margin of error to be no more than 0.03. This means that

$1.645\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\le 0.03\Leftrightarrow \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\le 0.018237$

$\Leftrightarrow \frac{\hat{p}(1-\hat{p})}{n}\le 0.000332591\Leftrightarrow n\ge \frac{\hat{p}(1-\hat{p})}{0.000332591}$

But $$\hat{p}$$ takes values between 0 and 1, so the maximum value of $$\hat{p}(1-\hat{p})$$ is achieved when $$\hat{p}=\frac{1}{2}$$. Therefore, the condition we need to satisfy is

$n\ge \frac{1}{4}\times \frac{1}{0.000332591}=751.674$

This means that the sample size should be at least $$n=752$$.