Q. 10.60 In this section, we introduced t... [FREE SOLUTION]

Chapter 10: Q. 10.60 (page 417)

In this section, we introduced the pooled t-test, which provides a method for comparing two population means. In deriving the pooled f-test, we stated that the variable
$z = \frac{({\hat{f}}_{1} - {\hat{x}}_{2}) - (渭_{1} - 渭_{2})}{蟽 \sqrt{(1 / n_{1}) + (1 / n_{2})}}$
cannot be used as a basis for the required test statistic because $蟽$ is unknown. Why can't that variable be used as a basis for the required test statistic?

Short Answer

Expert verified

$s_{p}$ integrates data about the variation of both samples into a single estimate of the population standard deviation's common value. As a result, $s_{p}$ stands for pooled standard deviation.

Step by step solution

Given Information

In deriving the pooled t- test, it is stated that the variable $z = \frac{({\overset{炉}{x}}_{1} - {\overset{炉}{x}}_{2}) - (渭_{1} - 渭_{2})}{蟽 \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}}$

Explanation

A test statistic is considered when performing a hypotheses test based on independent samples to compare the means of two populations with equal and unknown standard deviation.

The variable $z = \frac{({\overset{炉}{x}}_{1} - {\overset{炉}{x}}_{2}) - (渭_{1} - 渭_{2})}{蟽 \sqrt{\frac{1}{n_{1}} + \frac{1}{n}}}$ cannot be used as a basis for the required test.

Because $蟽$ is unknown,

In this case, the parameter $蟽$ represents the average standard deviation of two populations. Because the standard deviation of the majority of the population is unknown in general, the common standard deviation of the two populations is also unlikely to be known.

As a result, the individual sample variances $s_{1}^{2}$ and $s_{2}^{2}$ of two populations are taken and pooled by weighting them based on their sample size or degree of freedom, $n_{1}$ and $n_{2}$

$s_{p} = \sqrt{\frac{(n_{1} - 1) s_{1}^{2} + (n_{2} - 1) s_{2}^{2}}{n_{1} + n_{2} - 2}}$