Chapter 8.2, Problem 23P

Expand Your knowledge: Alternate Method for Confidence Intervals When a is unknown and the sample is of size $n\ge 30,$, there are two methods for computing confidence intervals for µ.

Method 1: use the Student’s t distribution with d.f. = n - 1.

This is the method used in the text. It is widely employed in statistical studies Also, most statistical software packages use this method.

Mrthod 2: When $n\ge 30,$ use the sample standard deviation s as an estimate for $\sigma$, and then use the standard normal distribution.

This method is based on the fact that for large samples, s is a fairly good approximation for $\sigma$. Also, for large n, the critical values for the Student’s t distribution approach those of the standard normal distribution.

Consider a random sample of size n = 31, with sample mean x = 45.2 and sample standard deviation s = 5.3.

(a) Compute 90%, 95%, and 99% confidence intervals for µ using Method 1 with a Student’s t distribution. Round endpoints to two digits after the decimal.

(b) Compute 90%, 95% and 99% confidence intervals for µ using Method 2 with the standard normal distribute. Use s as an estimate for $\sigma$. Round endpoints to two digits after the decimal.

(c) Compare intervals for the two methods. Would you say that confidence intervals using a Student’s t distribute are more conservative in the sense that they tend to be longer than intervals based on the standard normal distribute ?

(d) Repeal parts (a) through (c) for a sample of size n = 81. With increased sample size, do the two methods give respective confidence intervals that are more similar?

To determine

To find: The90%, 95%, and 99% confidence intervals for $\mu$ using Method1 with a Student’s t distribution.

Answer to Problem 23P

Solution:

The 90%, 95% and 99% confidence interval for $\mu$ are (43.58, 46.82), (43.26, 47.14), (42.58, 47.82) respectively.

Explanation of Solution

Calculation:

Let $\overline{x}$ represents the sample mean, i.e. $\overline{x}=45.2$, sample standard deviation $s=5.3$ and sample size $n=31$.

We have to find 90% confidence interval,

$\begin{array}{l}c=0.90\\ n=31\\ d.f.=n-1=31-1=30\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ t_{0.90}=1.697\\ s=5.3\\ E=t_c\frac{s}{\sqrt{n}}\\ E=1.697\frac{5.3}{\sqrt{31}}\\ E=1.6154\\ E\approx 1.62\end{array}$

90% confidence interval is

$\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 45.2-1.62<\mu <45.2+1.62\\ 43.58<\mu <46.82\end{array}$

The 90% confidence interval for $\mu$ is (43.58,46.82).

We have to find 95% confidence interval,

$\begin{array}{l}c=0.95\\ n=31\\ d.f.=n-1=31-1=30\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ t_{0.95}=2.042\\ s=5.3\\ E=t_c\frac{s}{\sqrt{n}}\\ E=2.042\frac{5.3}{\sqrt{31}}\\ E=1.944\\ E\approx 1.94\end{array}$

95% confidence interval is

$\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 45.2-1.94<\mu <45.2+1.94\\ 43.26<\mu <47.14\end{array}$

The 95% confidence interval for $\mu$ is (43.26, 47.14).

We have to find 99% confidence interval,

$\begin{array}{l}c=0.99\\ n=31\\ d.f.=n-1=31-1=30\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ t_{0.95}=2.750\\ s=5.3\\ E=t_c\frac{s}{\sqrt{n}}\\ E=2.750\frac{5.3}{\sqrt{31}}\\ E=2.6177\\ E\approx 2.62\end{array}$

99% confidence interval is

$\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 45.2-2.62<\mu <45.2+2.62\\ 42.58<\mu <47.82\end{array}$

The 99% confidence interval for $\mu$ is (42.58, 47.82).

To determine

To find: The 90%, 95%, and 99% confidence intervals for $\mu$ using Method 2 with the standard normal distribution.

Answer to Problem 23P

Solution:

The 90%, 95% and 99% confidence interval for $\mu$ are (43.63, 46.77), (43.33, 47.07), (42.74, 47.66) respectively.

Explanation of Solution

Calculation:

Let $\overline{x}$ represents the sample mean, i.e. $\overline{x}=45.2$, sample standard deviation $s=5.3$ and sample size $n=31$. Assume that $\sigma$ is unknown. Since sample size is greater than 30, hence we can use s as an estimate for $\sigma$.

We have to find 90% confidence interval,

$\begin{array}{l}c=0.90\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ z_{0.95}=1.645\\ s=5.3\\ n=31\\ E=z_c\frac{s}{\sqrt{n}}\\ E=1.645\frac{5.3}{\sqrt{31}}\\ E=1.5659\\ E\approx 1.57\end{array}$

90% confidence interval is

$\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 45.2-1.57<\mu <45.2+1.57\\ 43.63<\mu <46.77\end{array}$

The 90% confidence interval for $\mu$ is (43.63, 46.77).

We have to find 95% confidence interval,

$\begin{array}{l}c=0.95\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ z_{0.95}=1.96\\ s=5.3\\ n=31\\ E=z_c\frac{s}{\sqrt{n}}\\ E=1.96\frac{5.3}{\sqrt{31}}\\ E=1.866\\ E\approx 1.87\end{array}$

95% confidence interval is

$\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 45.2-1.87<\mu <45.2+1.87\\ 43.33<\mu <47.07\end{array}$

The 95% confidence interval for $\mu$ is (43.33, 47.07).

We have to find 99% confidence interval,

$\begin{array}{l}c=0.99\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ z_{0.99}=2.58\\ s=5.3\\ n=31\\ E=z_c\frac{s}{\sqrt{n}}\\ E=2.58\frac{5.3}{\sqrt{31}}\\ E=2.456\\ E\approx 2.46\end{array}$

99% confidence interval is

$\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 45.2-2.46<\mu <45.2+2.46\\ 42.74<\mu <47.66\end{array}$

The 99% confidence interval for $\mu$ is (42.74, 47.66).

To determine

Whether the confidenceintervals using a Student’s t distribution are more conservative.

Answer to Problem 23P

Solution:

Yes, the confidenceintervals using a Student’s t distribution are more conservative.

Explanation of Solution

Yes, the confidenceintervals using a Student’s t distribution is slightly longer than the intervals based on the standard normal distribution. Hence, the respective confidenceintervals using a Student’s t distribution are more conservative.

To determine

The 90%, 95%, and 99% confidence intervals for $\mu$ using Method 1 and Method 2.

Answer to Problem 23P

Solution:

The 90%, 95% and 99% confidence interval for $\mu$ using Method 1 are (44.22, 46.18), (44.03, 46.37), (43.65, 46.75) respectively. The 90%, 95% and 99% confidence interval for $\mu$ using Method 2 are (44.23, 46.17), (44.05, 46.35), (43.68, 46.72) respectively.

Explanation of Solution

Let $\overline{x}$ represents the sample mean, i.e. $\overline{x}=45.2$, sample standard deviation $s=5.3$ and sample size $n=81$.

Using Method 1 with a Student’s t distribution:

We have to find 90% confidence interval,

$\begin{array}{l}c=0.90\\ d.f.=n-1=81-1=80\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ t_{0.90}=1.664\\ E=t_c\frac{s}{\sqrt{n}}\\ E=1.664\frac{5.3}{\sqrt{81}}\\ E=0.9799\\ E\approx 0.98\end{array}$

90% confidence interval is

$\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 45.2-0.98<\mu <45.2+0.98\\ 44.22<\mu <46.18\end{array}$

The 90% confidence interval for $\mu$ is (44.22, 46.18).

We have to find 95% confidence interval,

$\begin{array}{l}c=0.95\\ d.f.=n-1=81-1=80\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ t_{0.95}=1.99\\ E=t_c\frac{s}{\sqrt{n}}\\ E=1.99\frac{5.3}{\sqrt{81}}\\ E=1.172\\ E\approx 1.17\end{array}$

95% confidence interval is

$\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 45.2-1.17<\mu <45.2+1.17\\ 44.03<\mu <46.37\end{array}$

The 95% confidence interval for $\mu$ is (44.03, 46.37).

We have to find 99% confidence interval,

$\begin{array}{l}c=0.99\\ d.f.=n-1=81-1=80\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ t_{0.95}=2.639\\ E=t_c\frac{s}{\sqrt{n}}\\ E=2.639\frac{5.3}{\sqrt{81}}\\ E=1.554\\ E\approx 1.55\end{array}$

99% confidence interval is

$\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 45.2-1.55<\mu <45.2+1.55\\ 43.65<\mu <46.75\end{array}$

The 99% confidence interval for $\mu$ is (43.65, 46.75).

Using Method 2 with the standard normal distribution:

We have to find 90% confidence interval,

$\begin{array}{l}c=0.90\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ z_{0.95}=1.645\\ s=5.3\\ n=81\\ E=z_c\frac{s}{\sqrt{n}}\\ E=1.645\frac{5.3}{\sqrt{81}}\\ E=0.9687\\ E\approx 0.97\end{array}$

90% confidence interval is

$\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 45.2-1.57<\mu <45.2+1.57\\ 44.23<\mu <46.17\end{array}$

The 90% confidence interval for $\mu$ is (44.23, 46.17).

We have to find 95% confidence interval,

$\begin{array}{l}c=0.95\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ z_{0.95}=1.96\\ s=5.3\\ n=81\\ E=z_c\frac{s}{\sqrt{n}}\\ E=1.96\frac{5.3}{\sqrt{81}}\\ E=1.154\\ E\approx 1.15\end{array}$

95% confidence interval is

$\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 45.2-1.15<\mu <45.2+1.15\\ 44.05<\mu <46.35\end{array}$

The 95% confidence interval for $\mu$ is (44.05, 46.35).

We have to find 99% confidence interval,

$\begin{array}{l}c=0.99\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ z_{0.99}=2.58\\ s=5.3\\ n=81\\ E=z_c\frac{s}{\sqrt{n}}\\ E=2.58\frac{5.3}{\sqrt{81}}\\ E=1.519\\ E\approx 1.52\end{array}$

99% confidence interval is

$\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 45.2-1.52<\mu <45.2+1.52\\ 43.68<\mu <46.72\end{array}$

The 99% confidence interval for $\mu$ is (43.68, 46.72).

The confidenceintervals using a Student’s t distribution is slightly longer than the intervals based on the standard normal distribution. As the sample size increases, the difference between the two methods is less pronounced. That means with increased sample size, the two methods give respective confidence intervals that are more similar.