Question

Confidence Interval with Known \(\sigma \). In Exercises 37 and 38, find the confidence interval using the known value of \(\sigma \).

Birth Weights of Boys Construct the confidence interval for Exercise 10 “Birth Weights of Boys,” assuming that \(\sigma \) is known to be 6.6 hg.

Short answer

The 95% confidence interval for the estimate mean is\(31.8\,{\rm{hg}} < \mu < 33.6\,{\rm{hg}}\).

Step-by-step solution

Given information

Refer to Exercise 10 for the summary statistics for randomly selected weights of boys as\(n\, = \,195,\bar x\, = 32.7\,{\rm{hg}}\).

The 95% confidence level with the known value of \(\sigma = 6.6\,\,{\rm{hg}}\).

Describe the confidence interval

A confidence interval is an estimate of the interval that may contain the true value of a population parameter. It is also known as an interval estimate.

The general formula for the confidence interval estimate of mean for known\(\sigma \)is as follows.

\({\rm{Confidence}}\,\,{\rm{interval}} = \left( {\bar x - {\rm{E}},\bar x + {\rm{E}}} \right)\,\,\,\,\,\,\,\,...\left( 1 \right)\)\(\)

Here, E is the margin of error, which is calculated as follows.

\({\rm{E}} = {z_{\frac{\alpha }{2}}} \times \frac{\sigma }{{\sqrt {\rm{n}} }}\).

Find the appropriate distribution

For a normally distributed population and randomly selected samples, the following are true.

If\(\sigma \)is known, the normal distribution is suitable to find the confidence interval.

If\(\sigma \)is unknown, the student’s t-distribution is suitable to find the confidence interval.

In this case,\(\sigma \)is known, and\({\rm{n}} = 205\),which is greater than 30.

Thus, normal distribution applies.

Find the critical value \({z_{\frac{\alpha }{2}}}\)

\({z_{\frac{\alpha }{2}}}\)is a z score that separates an area in the right tail of the standard normal distribution.

The confidence level 95% corresponds to\(\alpha = 0.05\,\,\,\,{\rm{and}}\,\,\,\,\frac{\alpha }{2} = 0.025\).

The value\({z_{\frac{\alpha }{2}}}\)hasthe cumulative area\(1 - \frac{\alpha }{2}\)to its left. .

Mathematically,

\(\begin{array}{c}P\left( {z < {z_{\frac{\alpha }{2}}}} \right) = 1 - \frac{\alpha }{2}\\ = 0.975\end{array}\)

From the standard normal table, the area of 0.975 is observed as the corresponding intersection of the row value 1.9 and column value 0.06, which implies that\({z_{\frac{\alpha }{2}}}\)is 1.96.

Find the margin of error

The margin of error is calculated as follows.

\(\begin{array}{c}{\rm{E}} = {z_{\frac{\alpha }{2}}} \times \frac{\sigma }{{\sqrt {\rm{n}} }}\\ = 1.96 \times \frac{{6.6}}{{\sqrt {195} }}\\ = 0.9264\end{array}\).

Find the confidence interval

The confidence interval is obtained by substituting the value of the margin of error in equation (1), as follows.

\(\begin{array}{c}{\rm{Confidence}}\,\,{\rm{Interval}} = \left( {\bar x - {\rm{E}},\bar x + {\rm{E}}} \right)\\ = \left( {32.7 - 0.9264\,\,,\,\,32.7 + 0.9264} \right)\\ = \left( {31.7736\,,\,\,33.6264} \right)\end{array}\)

Thus, the 95% confidence interval for the estimate mean is \(31.8\,{\rm{hg}} < \mu < 33.6\,{\rm{hg}}\).