Chapter 29.7, Problem 8RE

(a)

To determine

Delineate whether the question makes sense.

Answer to Problem 8RE

The question does not make sense.

Explanation of Solution

In this scenario, 50% of the women aged 16 years and older in Country U were employed in the year 1985 and 59% of the women 16 and older in the year 2005.

Based on the given information, the following values are known:

$\begin{array}{l}\text{Percentage }A=50\%\\ \text{Percentage }B=59\%\\ \text{Sample size }A=50,000\\ \text{Sample size }B=50,000\end{array}$

The question of statistically significant difference in percentages only makes sense when the sample is a probability sample.

On observing the given information, the sample is not known to be a probability sample, as the sample could contain all individuals in the population, and thus, the individuals were not selected from the population by chance. Therefore, this implies that the question does not make sense.

(b)

To determine

Delineate whether the information provided makes sense.

Answer to Problem 8RE

The information provided does not make sense.

Explanation of Solution

It is known from the results in Part (b) that the question does not make sense. In this context, it is known whether the samples are simple random samples.

This is not appropriate to answer the question as long as whether or not the samples are simple random samples.

(c)

To determine

State whether the difference in percentages is statistically significant if it is assumed that the Current Population Survey was based on independent simple random samples in each year of 50,000 women aged 16 years and older.

Answer to Problem 8RE

There is enough evidence that the difference in percentages is statistically significant.

Explanation of Solution

The box corresponding to 1985 data contains a ticket per woman, where the value on the ticket is a 1, when she is employed. Otherwise, the value on the ticket is 0.

The box corresponding to 2005 data contains a ticket per woman, where the value on the ticket is a 1, when she is employed. Otherwise, the value on the ticket is 0.

The test hypotheses are given below:

Let $\mu$ be the population mean.

Null hypothesis: $H_0:\mu =0$

That is, the percentages of the two boxes are the same.

Alternative hypothesis: $H_1:\mu \ne 0$

That is, the percentages of the two boxes are different.

The standard deviation A is given below:

$\begin{array}{c}\text{SE}A=\left(\text{Big number}-\text{Small number}\right)\sqrt{\left\{\begin{array}{l}\text{Fraction with big number}\times \\ \text{Fraction with small number}\end{array}\right\}}\\ =\left(1-0\right)\sqrt{0.5\times \left(1-0.5\right)}\\ =\left(1-0\right)\sqrt{0.5\times 0.5}\\ =0.5\end{array}$

The standard error of Box A is given below:

$\begin{array}{c}\text{SE}\text{Sum}A=\sqrt{\text{Sample size }}\times \text{SD }A\\ =\sqrt{50,000}\times 0.5\\ =111.8034\end{array}$

The standard error A of the average is the standard error of the sum divided by the number of draws as given below:

$\begin{array}{c}\text{SE avearge }A=\frac{\text{SE sum }A}{\text{Number of draws}}\times 100\%\\ =\frac{111.8034}{50,000}\times 100\%\\ =0.22\%\end{array}$

The standard deviation B is given below:

$\begin{array}{c}\text{SE}B=\left(\text{Big number}-\text{Small number}\right)\sqrt{\left\{\begin{array}{l}\text{Fraction with big number}\times \\ \text{Fraction with small number}\end{array}\right\}}\\ =\left(1-0\right)\sqrt{0.59\times \left(1-0.41\right)}\\ =\left(1-0\right)\sqrt{0.59\times 0.41}\\ =0.4918\end{array}$

The standard error of Box B is given below:

$\begin{array}{c}\text{SE}\text{Sum}B=\sqrt{\text{Sample size }}\times \text{SD }B\\ =\sqrt{50,000}\times 0.4918\\ =109.9698\end{array}$

The standard error B of the average is the standard error of the sum divided by the number of draws as given below:

$\begin{array}{c}\text{SE avearge }B=\frac{\text{SE sum }B}{\text{Number of draws}}\times 100\%\\ =\frac{109.9698}{50,000}\times 100\%\\ =0.22\%\end{array}$

The standard error for the difference is shown below:

$\begin{array}{c}\text{SE difference}=\sqrt{{\left(\text{SE first quantity}\right)}^2+\left({\left(\text{SE second quantity}\right)}^2\right)}\\ =\sqrt{{0.22}^2+{0.22}^2}\\ =0.31\%\end{array}$

The formula for test statistic is as follows:

$z=\frac{\text{Observed}-\text{Expected}}{\text{SE}}$

Known values:

$\begin{array}{c}\mu =0\\ \sigma =0.31\end{array}$

The observed value, x is given below:

$\begin{array}{c}x=\text{Percentage }A-\text{Percentage}B\\ =50\%-59\%\\ =-9\%\end{array}$

The z-score is obtained as given below:

$\begin{array}{c}z=\frac{-9-\text{0}}{\text{0}\text{.31}}\\ =-29.05\end{array}$

The P-value is given below:

$\begin{array}{c}P=P\left(Z<-29.05\text{ or }Z>29.05\right)\\ =1-\left[P\left(Z<29.05\right)-P\left(Z<-29.05\right)\right]\end{array}$

Using the Standard normal table, the value corresponding to $P\left(Z<-29.05\right)$ is 0 and $P\left(Z<29.05\right)$ is 1.

Remaining calculation:

$\begin{array}{c}P=1-\left[P\left(Z<29.05\right)-P\left(Z<-29.05\right)\right]\\ =1-\left[1-0\right]\\ =1-1\\ =0\end{array}$

Since the P-value is less than any of level of significance, it is unusual to obtain a difference in sample percentages of 9% when there is no difference in the population percentages, and thus, the difference does not appear to be due to chance variation.

Therefore, this implies that the difference in percentages is statistically significant.