In each of the following instances, determine whether you would use a z- or t-statistic (or neither) to form a 90%confidence interval and then state the appropriate z- ort-statistic value for the confidence interval.

a. Random sample of size n = 32 from a normal distribution with a population mean of 60 and population standard deviation of 4.

b. Random sample of size n = 108 from an unknown population.

c. Random sample of size n = 12 from a normal distribution witha sample mean of 83 and sample standard deviation of 2.

d. Random sample of size n = 24 from a normal distribution withan unknown mean and sample standard deviation of 3.

A Z-test and T-test are two statistical hypothesis techniques that help evaluate the values of critical points at a level of significance in order to accept or reject the hypothesis formulated.

• If the population standard deviation is not known or if the sample size is less than 30, need to use the T-Test.
• If the population standard deviation is known and when the sample size is greater than 30 and comes from a normal distribution, need to use Z-test.

The value of Z-test is:

$\begin{array}{rcl}Z& =& \frac{X-\mu }{\frac{\sigma }{\sqrt{n}}}\\ & =& \frac{X-60}{\frac{4}{\sqrt{32}}}\\ & & \end{array}$

For a 90% confidence interval the Z-value will be:1.645

Since the population is unknown, we use the z-test as n=108, which is greater than 30. This implies sample size is quite large.

For a 90% confidence interval the Z-value will be:1.645

Since the population is normal, n=12 which is less than 30 and sample standard deviation is 2, given,$n=12, \overline{x}=83, s=2$ . This implies one have to use t-test.

The value of T-test is:

$\begin{array}{rcl}t& =& \frac{\overline{X}-\mu }{\frac{s}{\sqrt{n}}}\\ & =& \frac{83-\mu }{\frac{2}{\sqrt{12}}}\\ & & \end{array}$

For a 90% confidence interval, the t-value will depend on parameter and degree of freedom. Since here n is 12, the degree of freedom is n-1, which is 12-1=11.

Therefore, t value for 11 degrees of freedom for 0.9 level of significance is 1.363.

Since the population is normal, n=24 which is less than 30 and sample standard deviation is 3, we have, $n=24, s=3$. This implies onee have to use t-test.

The value of T-test is:

$$\begin{gathered} t=\frac{\overline{X}-\mu }{\frac{s}{\sqrt{n}}} \\ =\frac{\overline{X}-\mu }{\frac{3}{\sqrt{24}}} \end{gathered}$$

For a 90% confidence interval, the t-value will depend on parameter and degree of freedom. Since here n is 24, the degree of freedom is n-1, which is 24-1=23.

Therefore, t value for 23 degrees of freedom for 0.9 level of significance is 1.319.