Confidence Intervals for Normal Variables

In the last section we considered point estimates   –   that is, estimates of a specific value.   In many problems of estimation it is instead preferable to have an interval estimate that contains the accuracy of the estimate.   To give such, we will use a method commonly known as the method of confidence intervals.   We introduce it here with a specific example.

Example: Find a   $90\%$   confidence interval for the mean of a normal random variable

The interval   $I=(-1.645, 1.645)$   has the property that   $\displaystyle{ P\left( Z_{0,1} \in I\right) =0.90 }\;$ .   We say that this interval is a     $90\%$   confidence interval for the standard normal random variable   $\displaystyle{ Z_{0,1} }\;$ .

Confidence Intervals Symmetric About the Mean: The General Plan

Given a normal random variable   $\displaystyle{ Z_{\mu,\sigma} }\,$ , with   $\mu$   and   $\sigma$   is known, we say   $[\mu -a, \mu +a]$   is an interval symmetric about the mean   $\mu\;$ .   If the probability   $\displaystyle{ P( Z_{\mu,\sigma} \in [\mu -a, \mu +a] ) =B }\,$ , we say that   $[\mu -a, \mu +a]$   is a symmetric interval about the mean of probability   $B\,$ , or a   $100\cdot B \%$   symmetric interval about the mean.   We will often say that   $[\mu -a, \mu +a]$   is a   $100\cdot B \%$   symmetric about the mean confidence interval for   $\displaystyle{ Z_{\mu,\sigma} }\,$ , as we know that   $100\cdot B \%$   of the probability of   $\displaystyle{ Z_{\mu,\sigma} }$   resides in this interval.

Thus, for example, $\, [-1.645, 1.645]$   was observed above to be a   $90\%$   confidence interval about the mean for a standard normal random variable.   We may also check that   $[-1.282, 1.282]$   is an   $80\%$   confidence interval about the mean for a standard normal random variable, and   $[0.065, 9.935]$   is a   $90\%$   confidence interval about the mean for a   $\displaystyle{ Z_{5,3} }$   normal random variable.

Example:   Find   $95\%$   confidence intervals (i) about the mean, (ii) one-sided to the left, and (iii) one-sided to the right for a standard normal random variable.

  1. To find the   $95\%$   confidence interval about the mean ( $\, =0\,$ ) for a standard normal distribution, we wish to find value   $m$   so that   $\displaystyle{ P(Z_{0,1}\in (-m,m)) =.95 }\;$ .   Using the symmetry of both the confidence interval and of   $\displaystyle{ Z_{0,1} }\,$ , we know that   $\displaystyle{ P(Z_{0,1} \le -m) =P(Z_{0,1} \ge m) =.025 }\;$ .   Thus   $\displaystyle{ P(Z_{0,1}\in (-\infty,m)) =.975 }\,$ , and we can find   $m$   in a spreadsheet with the command “=NORMSINV(0.975)” .   We get the value   $m =1.96\;$ .

  2. To find the   $95\%$   confidence interval one-sided to the left for a standard normal distribution, we wish to find value   $m$   so that   $\displaystyle{ P(Z_{0,1}\in (-\infty,m)) =.95 }\;$ .   Again we can find   $m$   in a spreadsheet with the command “=NORMSINV(0.95)” .   We get the value   $m =1.64\;$ .

  3. Finally, to find the   $95\%$   confidence interval one-sided to the right for a standard normal distribution, we use the symmetry of   $\displaystyle{ Z_{0,1} }$   to observe that   $\displaystyle{ P(Z_{0,1}\in (m,\infty)) =P(Z_{0,1}\in (-\infty,-m)) =.95 }\;$ .   Thus we find   $-m$   using the previous item in this example, and from that get   $m\;$ .   We see that   $\displaystyle{ P(Z_{0,1}\in (-\infty,-m)) =.95 }$   for   $-m =1.64\,$ , so   $m =-1.64\;$ .

We can do the same thing if we have a non-standard random variable   $\displaystyle{ Z_{\mu,\sigma} }\;$ .   We use the fact that   $\displaystyle{ P\left( Z_{\mu,\sigma} \lt A \right) =P\left( Z_{0,1} \lt \sigma\ A +\mu \right) }\;$ .   Thus, for example, since   $(-1.64, 1.64)$   is a   $90\%$   confidence interval for a standard normal variable, $\, (\mu -1.64\cdot\sigma, \mu +1.64\cdot\sigma)$   is a   $90\%$   confidence interval for   $\displaystyle{ Z_{\mu,\sigma} }\;$ .

Example:   Find   $95\%$   confidence intervals (i) about the mean, (ii) one-sided to the left, and (iii) one-sided to the right for a   $\displaystyle{ Z_{5,3} }$   random variable.

  1. To find the   $95\%$   confidence interval about the mean ( $\, =5\,$ ) for a   $\displaystyle{ Z_{5,3} }$   normal distribution, we wish to find value   $m$   so that   $\displaystyle{ P(Z_{5,3}\in (5-m,5+m)) =.95 }\;$ .   But this is the probability that   $\displaystyle{ P(Z_{0,3}\in (-m,+m)) =.95 }\,$ , or   $\displaystyle{ P(Z_{0,1}\in \left( -\textstyle{\frac{m}{3}},\textstyle{\frac{m}{3}}\right)) =.95 }\;$ .   But we have already seen that   $\displaystyle{ P(Z_{0,1}\in \left( -1.960, 1.960\right)) =.95 }$ .   Thus we know that   $\frac{m}{3} =1.960\,$ ,   or   $m =5.880\;$ .   The desired interval is   $(5-5.880,5+5.880) =(-0.880, 10.880)\;$ .   The bounds can be obtained using spreadsheet commands with “=NORMINV(0.025,5,3)” and “=NORMINV(0.975,5,3)” .

  2. To find the   $95\%$   confidence interval one-sided to the left for a   $\displaystyle{ Z_{5,3} }$   normal distribution, we wish to find value   $m$   so that   $\displaystyle{ P(Z_{5,3}\in (-\infty,5+m)) =.95 }\;$ .   But this is the probability that   $\displaystyle{ P(Z_{0,3}\in (-\infty,m)) =.95 }\,$ , or   $\displaystyle{ P(Z_{0,1}\in \left( -\infty,\textstyle{\frac{m}{3}}\right)) =.95 }\;$ .   We have already found that   $\displaystyle{ P(Z_{0,1}\in ( -\infty,1.645) =.95 }\;$ .   Thus we know that   $\frac{m}{3} =1.645\,$ ,   or   $m =4.935\;$ .   The desired interval is   $(-\infty, 5+4.935) =(-\infty, 9.935)\;$ .   Again, this can be obtained using spreadsheet commands with “=NORMINV(0.95,5,3)” .

  3. Finally, to find the   $95\%$   confidence interval one-sided to the right for a   $\displaystyle{ Z_{5,3} }$   normal distribution, we wish to find value   $m$   so that   $\displaystyle{ P(Z_{5,3}\in (5-m, \infty)) =.95 }\;$ .   But this is the probability that   $\displaystyle{ P(Z_{0,3}\in (-m, \infty)) =.95 }\,$ , or   $\displaystyle{ P(Z_{0,1}\in \left( -\textstyle{\frac{m}{3}}, \infty\right)) =.95 }\;$ .   We have already found that   $\displaystyle{ P(Z_{0,1}\in ( -1.645, \infty) =.95 }\;$ .   Thus we know that   $-\frac{m}{3} =-1.645\,$ ,   or   $m =-4.935\;$ .   The desired interval is   $(5-4.935, \infty) =(.065, \infty)\;$ .

Note:   Since the sum of two independent normal random variables is a normal random variable, and since the negative of a normal random variable is also a normal random variable, we see that the difference of two independent random normal variables is a can use this to understand where the sum or difference of independent random variables takes values.

Example: Find symmetric about the mean   $90\%$   confidence intervals for the sum and difference of two independent normal random variables.

Let   $\displaystyle{ X_1 }$   and   $\displaystyle{ X_2 }$   be independent normal random variables with mean   $18$   and   $23\,$ , and standard deviations   $3$   and   $4\,$ , respectively.   Then   $\displaystyle{ X_1 +X_2 }$   and   $\displaystyle{ X_1 -X_2 }$   are normal random variables with means   $18 +23 =41$   and   $18 -23 =-5\,$ , and both with standard deviation   $\displaystyle{ \sqrt{3^2 +4^2} =5 }\;$ .

A symmetric about the mean   $90\%$   confidence interval for   $\displaystyle{ X_1 +X_2 }$   will be
$$ [(18+23) -1.645\cdot 5, (18+23) +1.645\cdot 5] =[32.776, 49.224] \;\text{.} $$
Similarly, a symmetric about the mean   $90\%$   confidence interval for   $\displaystyle{ X_1 -X_2 }$   will be
$$ [(18-23) -1.645\cdot 5, (18-23) +1.645\cdot 5] =[-13.224, 3.224] \;\text{.} $$