Identify the assumption for using the two means $z$test and the two mean $z$interval procedure that renders those procedures generally impractical.

Given In the question that Two means $z$tests and two means $z$interval Procedure are used for comparing population mean. we have to calculate the assumption for using two mean $z$test and two mean $z$interval procedure.

he two-means$z$test and the two-means $z$-interval procedure require the presence of two normally distributed but independent populations with known standard deviations ${\sigma }_{1}and{\sigma }_2$.

The hypotheses test formula is as follows:

$z=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_1^2}{n_1}}}$

Where $\overline{x_1}and\overline{x_2}$are the means of two sample.

${\mu }_1-{\mu }_2$-Hypothesized difference between the population mean.

${\sigma }_{1}and{\sigma }_2$-Standard deviation

$n_{1}and{n}_2$-size of the two samples.

Null hypothesis: $H_0:{\mu }_1={\mu }_{2}OR{H}_0:{\mu }_1-{\mu }_2=0$

Alternate hypothesis : $H_0:{\mu }_1\ne {\mu }_{2}OR{H}_0:{\mu }_1-{\mu }_2\ne 0$

This method is works only if standard deviation ${\sigma }_{1}and{\sigma }_2$are known.