**1. i. Compare Z distribution and Normal Distributions. ii. What are the properties of the t distribution? iii. Find the values for each. a. t š¼/2 and n = 15 for the 98% confidence interval for the mean b. t š¼/2 and n = 10 for the 90% confidence interval for the mean**

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### Solution:

1 (i):

A normal distribution is determined by two parameters the mean and the variance. Often in statistics we refer to an arbitrary normal distribution as we would in the case where we are collecting data from a normal distribution in order to estimate these parameters. Now the standard normal distribution is a specific distribution with mean 0 and variance 1

. This is the distribution that is used to construct tables of the normal distribution.

Conveniently ifĀ *XĀ *has the a normal distribution with meanĀ *mĀ *and varianceĀ *s*2Ā then if we define$Z=\dfrac{X-m}s$

thenĀ *ZĀ *has the standard normal distribution. So for any specific normal distribution we can calculate probabilities of the formĀ *P*[*a*<*X*<*b*]

from the tables forĀ *Z.*Ā This is because we can writeĀ *X*=*sZ*+*m*.

So the standard normal plays a special role with respect to the general family of normal distributions.

(ii):

The t distribution has the following properties:

The mean of the distribution is equal to 0 .

The variance is equal to v / ( v - 2 ), where v is the degrees of freedom (see last section) and v > 2. The variance is always greater than 1, although it is close to 1 when there are many degrees of freedom.

(iii):

*(a) tĀ **š¼*/2Ā andĀ *n =*Ā 15 for the 98% confidence interval for the mean

$\alpha=100\%-98\%=2\% \\\alpha/2=1\%=0.01$

Using t-table value:

$t_{0.01,14}=2.624$

*(b) tĀ **š¼*/2Ā andĀ *n =*Ā 10 for the 90% confidence interval for the meanĀ

$\alpha=100\%-90\%=10\% \\\alpha/2=5\%=0.05$

Using t-table value:

$t_{0.05,9}=1.833$