Question

Refer to Procedure $8.1.$

a. Explain in detail the assumptions required for using the $z-$interval procedure.

b. How important is the normality assumption? Explain your answer.

Short answer

Part (a) If $\sigma$is not known we can not determine the confidence limits.

Part (b) The assumption is critical.

Step-by-step solution

Part (a) Step 1: Given information

For any normally distributed variable x with mean ${\mu }_x$and standard deviation${\sigma }_x$ probability that the value of the variable will be within the interval.

Part (a) Step 2: Explanation

The following are the assumptions:

i. simple random sample

ii. normal population or large sample

iii. $\sigma$Known.

Assumption (i) simple random sample In simple random sampling, each sample of a fixed size has the same chance of being drawn from the population. Thus, simple random sampling assures that we are not biased toward any particular sample or sample mean value, which could alter the $\mu$confidence interval.

Assumption (ii) big sample or population

To ensure that the sample mean is normally or substantially normally distributed, this assumption is required. Because the sample mean follows the normal distribution of the population variable, sample means are approximately normally distributed when the sample size is large.

Part (a) Step 3: Calculation

Assumption (iii) $\sigma$Known

If $\sigma$is unknown, the confidence interval of the population mean cannot be calculated using the $z-$interval technique.

Because the sample mean follows a normal (or nearly normal) distribution with a mean of $\mu$and a standard deviation of $\frac{\sigma }{\sqrt{n}}$

Where $\sigma$is known and the $100(1-\alpha )\%$Confidence interval of $\mu$ using $z-$interval procedure is given by $\left(\overline{x}-z_{\frac{a}{2}}\times \frac{\sigma }{\sqrt{n}},\overline{x}+z_{\frac{a}{2}}\times \frac{\sigma }{\sqrt{n}}\right)$

As a result, if $\sigma$ is unknown, the confidence bounds cannot be determined.

Part (b) Step 1: Calculation

Because the one means $z-$interval process of obtaining the $100(1-\alpha )\%$confidence interval of the population mean is based on the notion that given a normally distributed variable $X$with mean ${\mu }_x$and S.D.${\sigma }_x$ the normality assumption is highly significant.
The probability (the chance) that the observed value of $X$will i.e in the interval

$P\left({\mu }_x-z_{\frac{a}{2}}{\sigma }_x<X<{\mu }_x+z_{\frac{\alpha }{2}}{\sigma }_x\right)=1-\alpha$

So, in order to apply the $z-$interval technique to get $100(1-\alpha )\%\mathrm{Cl}$ at $\mu$ $\overline{x}$ must be normally (or almost normally) distributed. As a result, the assumption is critical.