Question

Labor Union Support. Labor Day was created by the U.S. labor movement over $100$years ago. It was subsequently adopted by most states as an official holiday. In a poll by Gallup, $1003$randomly selected adults were asked whether they approve of labor unions; $65\%$said yes.

(a) In $1936$, about $72\%$of Americans approved of labor unions. At the $5\%$significance level, do the data provide sufficient evidence to conclude that the percentage of Americans who approve of labor unions now has decreased since $1936$?

(b) In $1963$, roughly $67\%$of Americans approved of labor unions. At the $5\%$significance level, do the data provide sufficient evidence to conclude that the percentage of Americans who approve of labor unions now has decreased since $1963$?

Short answer

(a) At the $5\%$level, the test results are statistically significant.

As a result, the data is adequate to establish that the percentage of Americans who favour of labour unions has fallen since $1936$.

(b) At the $5\%$level, the test results are not statistically significant.

As a result, the data does not support the conclusion that the number of Americans who favour of labour unions has fallen since $1963.$

Step-by-step solution

Part (a) Step 1: Given information

The significance level is $5\%$

The number of randomly selected adults is $n=1003$

The number of Americans approved of labor unions is$p_0=0.72$

The approval of labor unions $\hat{p}=0.65$

Part (a) Step 2: Explanation

Using the given data

Calculate the value of $n{p}_0$

$n{p}_0=\left(1003\right)(0.72)$

$=722.16$

Calculate the value of $n\left(1-p_0\right)$

$n\left(1-p_0\right)=\left(1003\right)(1-0.72)$

$=280.84$

Both the values are greater than $5$. So one proportion $z-$test is appropriate to use.

The null hypothesis is

$H_0:p_0=0.72$

The alternate hypothesis is

$H_0:p_0<0.72$

The $z$value is calculated as

$z=\frac{0.65-0.72}{\sqrt{\frac{0.72(1-0.72)}{1003}}}$

$=\frac{-0.07}{0.014}$

$=-4.937$

When $\alpha =0.05$the tail is left tailed

The critical value of $z$from the standard table is

$z_{0.05}=-1.645$

The test statistic is in the negative range. As a result, the hypothesis $H_0$is ruled out.

At the $5\%$level, the test results are statistically significant.

As a result, the data is adequate to establish that the percentage of Americans who favour of labour unions has fallen since $1936$.

Part (b) Step 1: Given information

The significance level is$5\%$

The number of randomly selected adults is$1003$

The number of Americans approved of labor unions is$p_0=0.72$

The approval of labor unions$\hat{p}=0.65$

Part (b) Step 2: Explanation

Calculate the value of$n{p}_0$

$n{p}_0=\left(1003\right)(0.65)$

$=672.01$

Calculate the value of$n\left(1-p_0\right)$

$n\left(1-p_0\right)=\left(1003\right)(1-0.67)$

$=330.99$

Both the values are greater than $5$. So one proportion $z-$test is appropriate to use.

The null hypothesis is

$H_0:p_0=0.67$

The alternate hypothesis is

$H_0:p_0<0.67$

The expression for $z$is

$z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}$

$z=\frac{0.65-0.67}{\sqrt{\frac{0.67(1-0.67)}{1003}}}$

$=\frac{-0.02}{0.015}$

$=-1.347$

Also, $\alpha =0.05$, the tail is left tailed

The critical value of $z$from the standard table is

$z_{0.05}=-1.645$

The test statistic is within acceptable bounds. As a result, the hypothesis $H_0$is not ruled out.

At the $5\%$level, the test results are not statistically significant.

As a result, the data does not support the conclusion that the number of Americans who favour of labour unions has fallen since $1963$.