Chapter 6, Problem 6.106LM

In each of the following instances determine whether you would use a z- or t-statistic (or neither) to form a 95% confidence interval and then look up the appropriate z- or t-value

1. a. Random sample of size n = 23 from a normal distribution with unknown mean μ and standard deviation σ
2. b. Random sample of size n − 135 from a normal distribution with unknown mean μ and standard deviation σ
3. c. Random sample of size n − 10 from a normal distribution with unknown mean  μ and standard deviation s − 5
4. d. Random sample of size n − 73 from a distribution about which nothing is known
5. e. Random sample of size n = 12 from a distribution about which nothing is known

To determine

To check and find: Whether z-or t-statistic is appropriate for finding the 95% confidence interval and the appropriate z- or t-value for the given information.

Answer to Problem 6.106LM

The t-statistic is appropriate for finding the 95% confidence interval.

The t-value corresponding to the given information is 2.074.

Explanation of Solution

Given info:

The given value is $n=23$. The distribution is normal with unknown mean and standard deviation and the level of confidence is 95%.

Calculation:

Conditions for confidence interval of $\mu$ when using z-statistic:

(i) Sample size is large. That is, $n\ge 30$.

(ii) If the population is of any distribution, then z-statistic can be used for confidence interval.

• If $\sigma$ is known, then it can be used.
• If $\sigma$ is unknown, then the estimate of $\sigma$ (s) can be used in the confidence interval.

Conditions for confidence interval of $\mu$ when using t-statistic:

(i) Sample size is less. That is, $n<30$.

(ii) If the population is a normal distribution, then t-statistic can be used for confidence interval.

• If $\sigma$ is known, then it can be used.
• If $\sigma$ is unknown, then the estimate of $\sigma$ (s) can be used in the confidence interval.

Here, the sample size is 23, which is less than 30. The mean and standard deviation are unknown.

Hence, the t-statistic can be used to find the confidence interval.

Critical value:

For 95% confidence level:

$\begin{array}{c}1-\alpha =1-0.95\\ \alpha =0.05\\ \frac{\alpha }{2}=\frac{0.05}{2}\\ =0.025\end{array}$

Degrees of freedom:

$\begin{array}{c}df=n-1\\ =23-1\\ =22\end{array}$

From “Table III of the t distribution in Appendix D”, the critical value is 2.074 for the degrees of freedom of 22.

To determine

To check and find: Whether z- or t-statistic is appropriate for finding the 95% confidence interval and the appropriate z- or t-value for the given information.

Answer to Problem 6.106LM

The z-statistic is appropriate for finding the 95% confidence interval.

The z-value corresponding to the given information is 1.96.

Explanation of Solution

Given info:

The value of $n=135$. The distribution is normal with unknown mean and standard deviation and the level of confidence is 95%.

Calculation:

Here, the sample size 135, which is greater than 30. Moreover, the mean and standard deviation are unknown. Hence, the population’s standard deviation can be estimated by the sample standard deviation.

Thus, the z-statistic can be used for finding the confidence interval.

Critical value:

For 95% confidence level:

$\begin{array}{c}1-\alpha =1-0.95\\ \alpha =0.05\\ \frac{\alpha }{2}=\frac{0.05}{2}\\ =0.025\end{array}$

Hence, the cumulative area to the left is as follows:

$\begin{array}{c}\text{Area to the left}=1-\text{Area to the right}\\ =1-0.025\\ =0.975\end{array}$

From “Table II of the standard normal distribution in Appendix D”, the critical value is 1.96.

To determine

To check and find: Whether z- or t-statistic is appropriate for finding the 95% confidence interval and the appropriate z- or t-value for the given information.

Answer to Problem 6.106LM

The t-statistic is appropriate for finding the 95% confidence interval.

The t-value corresponding to the given information is 2.262.

Explanation of Solution

Given info:

The given value are $n=10$ and $s=5$. The distribution is normal with unknown mean and the level of confidence is 95%.

Calculation:

Here, the sample is from a normal distribution with a size is 10, which is less than 30, but the standard deviation is given.

Therefore, the t-statistic can be used for finding the confidence interval.

Critical value:

For 95% confidence level:

$\begin{array}{c}1-\alpha =1-0.95\\ \alpha =0.05\\ \frac{\alpha }{2}=\frac{0.05}{2}\\ =0.025\end{array}$

From “Table III of the t distribution in Appendix D”, the critical value is 2.262.

To determine

To check and find: Whether z- or t-statistic is appropriate for finding the 95% confidence interval and the appropriate z- or t-value for the given information.

Answer to Problem 6.106LM

The z-statistic is appropriate for finding the 95% confidence interval.

The z-value corresponding to the given information is 1.96.

Explanation of Solution

Given info:

The given value is $n=73$ and no information is given about distribution.

Calculation:

Here, the distribution is unknown with a sample size of 73, which is larger than 30 and the population’s standard deviation can be estimated by the sample standard deviation.

Therefore, the z-statistic can be used for finding the confidence interval.

Critical value:

For 95% confidence level:

$\begin{array}{c}1-\alpha =1-0.95\\ \alpha =0.05\\ \frac{\alpha }{2}=\frac{0.05}{2}\\ =0.025\end{array}$

Hence, the cumulative area to the left is as follows:

$\begin{array}{c}\text{Area to the left}=1-\text{Area to the right}\\ =1-0.025\\ =0.975\end{array}$

From “Table II of the standard normal distribution in Appendix D”, the critical value is 1.96.

To determine

To check and find: Whether z- or t-statistic is appropriate for finding the 95% confidence interval and the appropriate z- or t-value for the given information.

Answer to Problem 6.106LM

z or t cannot be used.

Explanation of Solution

Given info:

The given value is $n=12$ and no information is given about the distribution.

Calculation:

Here, the sample size is 10, which is less than 30, and no information is provided about the distribution.

Also, the above conditions are not satisfied.

Therefore, the z- or t-statistic cannot be used for finding the confidence interval.