Question

Losing weight A Gallup poll found that 59% of the people in its sample said “Yes” when asked, “Would you like to lose weight?” Gallup announced: “For results based on the total sample of national adults, one can say with 95% confidence that the margin of (sampling) error is ±3 $\pm 3$percentage points.”12 Based on the confidence interval, is there convincing evidence that the true proportion of U.S. adults who would say they want to lose weight differs from 0.55? Explain your reasoning

Short answer

The required answer is :

Yes, there are sufficient evidence

Step-by-step solution

Given information

Sample proportion $\left(\hat{p}\right)=59\%=0.59$

Margin of error$\left(E\right)=\pm 3\%=\pm 0.03$

The objective is to find out the confidence interval 

The formula for calculating the confidence interval for a population proportion is as follows:

Confidence interval $=\hat{p}\pm E$

The $95\%$confidence interval can be calculated as follows:

Confidence interval :

$$\begin{gathered} =\hat{p}\pm E \\ =0.59\pm 0.03 \\ =(0.56,0.62) \end{gathered}$$

Therefore, the confidence interval is$(0.56,0.62)$

The objective is to find whether there are sufficient evidence to conclude that the proportion of U.S adults who are saying that they are interesting in losing the weight is different from 0.55 . 

Because $0.55$does not fall within the above-mentioned confidence interval. Thus, there is sufficient evidence to conclude that the proportion of US adults who say they want to lose weight is greater than $0.55$