Question

11.91 Economic Stimulus. In a national poll, $1053$ U.S. adults were asked, "As you may know, Congress is considering a new economic stimulus package of at least $800$ billion dollars. Do you favor or oppose Congress passing this legislation?" Of those sampled, $548$ favored passage.
a. At the $5\%$ significance level, do the data provide sufficient evidence to conclude that a majority (more than $50\%$ ) of U.S. adults favored passage?
b. The headline on the website featuring the survey read, "In U.S., Slim Majority Supports Economic Stimulus Plan." In view of your result from part (a), discuss why the headline might be misleading.
c. How could the headline be made more precise?

Short answer

(a) At the $5\%$ level, the test results are not statistically significant. The data does not support the conclusion that the majority of Americans approved passage.
(b) At the $5\%$significance level, from part (a). "In the United States, a Slender Majority Supports Economic Stimulus Plan," So, cannot conclude.
(c) "Slim Majority ($50\%$) of Respondents Supports Economic Stimulus Plan,"

Step-by-step solution

Part (a) Step 1: Given information

To find the data provide sufficient evidence to conclude that majority of U.S favored passages at the $5\%$significance level.

Part (a) Step 2: Explanation

Let, the significance level is $5\%,n=1053,p_0=0.5$and $x=548$.
Determine the value of $n{p}_0$.
$n{p}_0=\left(1053\right)(0.5)$

$=526.5$

The values $n{p}_0$and $n(1-p_0)$are both greater than $5$.

As a result, only one proportion $z$test should be used.

The null hypothesis: $H_0:p_0=0.5$

The alternate hypothesis: $H_{\mathrm{a}}:p_0>0.5$

Part (a) Step 3: Explanation

Determine the sample proportion as:
$\begin{array}{r}\widehat{p}=\frac{x}{n}\end{array}$
$=\frac{548}{1053}$
$=0.5204$
For $z$, define the expression.
$z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}$
$\begin{array}{r}=\frac{0.5204-0.5}{\sqrt{\frac{0.5(1-0.5)}{1053}}}\end{array}$
$=\frac{0.0204}{0.0154}$
$=1.324$
Also,
$\alpha =0.05$
The test is left tailed.

Part (a) Step 4: Explanation

The crucial value of $z$from the usual table for $\alpha =0.05$ is,
$z_{0.05}=-1.645$
The test statistic is within acceptable bounds.
As a result, the hypothesis $H_0$ is not rejected.
At the $5\%$ level, the test results are not statistically significant.
As a result, the data does not support the conclusion that the majority of Americans approved passage.

Part (b) Step 1: Given information

The headline on the website featuring the survey read, "In U.S., Slim Majority Supports Economic Stimulus Plan."

Part (b) Step 2: Explanation

Since, the headline is "In U.S Slim Majority Supports Economic Stimulus Plan."
In the United States, a slim majority of people support the economic stimulus plan.
The test statistic is within acceptable bounds.
As a result, the hypothesis $H_0$ is not ruled out.
At the $5\%$ level, the test result is not statistically significant.
As a result, it is impossible to conclude that more than $50\%$of US adults favored law approval.
As a result, the rationale for the misleading headline is stated above.

Part (c) Step 1: Given information

To find that the headline be made more precise.

Part (c) Step 2: Explanation

"In the United States, a Slender Majority Supports Economic Stimulus Plan," the headline reads.
In the United States, a slim majority of people support the economic stimulus plan.
When more than $50\%$ of US people approved passage of the Act, a more accurate headline could be used.