Question

Body Mass Index. Body mass index (BMI) is a measure of body fit based on height and weight. According to the document Dietary Guidelines for Americans, published by the U.S. Department of Agriculture and the U.S. Department of Health and Human Services, for adults, a BMI of greater than 25 indicates an above healthy weight (i.e., overweight or obese), Oct 750 randomly selected adults whose highest degree is a bachelor's, 386 have an above healthy weight; and of 500 randomly selected adults with a graduate degree, 237 have an above healthy weight.

a. What assumptions are required for using the two-proportions z-lest here?

b. Apply the two-proportions z-test to determine, at the 5% significance level, whether the percentage of adults who have an above healthy weight is greater for those whose highest degree is a bachelor's than for those with a graduate degree.

Short answer

a) The assumptions of the two proportions are satisfied.

b) There is no evidence to infer that the percentage of adults who have an above healthy weight is greater for those whose highest degree is a bachelor's than for those with a graduate degree.

Step-by-step solution

Part(a) Step 1: Given Information

The given values are,

$$\begin{gathered} x_1=386, \\ {n}_1=750, \\ {x}_2=237, \\ {n}_2=500, \\ \alpha =0.05 \end{gathered}$$

Part(a) Step 2: Explanation

The formula for ${\tilde{p}}_1$is given by,

${\tilde{p}}_1=\frac{x_1}{n_1}$

The formula for ${\tilde{p}}_2$is given by,

${\tilde{p}}_2=\frac{x_2}{n_2}$

The formula for$z$is given by,

$z=\frac{{\tilde{p}}_1-{\tilde{p}}_2}{\sqrt{{\tilde{p}}_p\left(1-{\tilde{p}}_p\right)}\sqrt{\frac{1}{n_1}+\frac{\Gamma }{m_2}}}$

The formula for${\tilde{p}}_p$is given by,

${\tilde{p}}_p=\frac{x_1+x_2}{n_1+n_2}$

The value of$n_1-x_1$is calculated as,

$$\begin{gathered} n_1-x_1=2701-35 \\ =2666 \end{gathered}$$

The value of$n_2-x_2$is calculated as,

$$\begin{gathered} n_2-x_2=2052-47 \\ =2005 \end{gathered}$$

The values $x_1,\left(n_1-x_1\right),x_2$, and $\left(n_2-x_2\right)$are all greater than 5 .

The value of${\tilde{p}}_1$is calculated as,

$$\begin{gathered} {\tilde{p}}_1=\frac{x_1}{n_1} \\ =\frac{386}{750} \\ =0.514\backslash \end{gathered}$$

The value of ${\tilde{p}}_2$is calculated as,

$$\begin{gathered} {\tilde{p}}_2=\frac{x_2}{n_2} \\ =\frac{237}{500} \\ =0.474 \end{gathered}$$

The value of ${\tilde{p}}_p$is calculated as,

$$\begin{gathered} {\tilde{p}}_p=\frac{x_1+x_2}{n_1+n_2} \\ =\frac{386+237}{750+500} \\ =0.4984 \end{gathered}$$

Assumptions:

The selected sample should be a simple random sample from two populations.

The samples are independent of one another.$x_1,n_1-x_1,x_2$and $n_2-x_2$are all at least 5.

The two samples are random samples, the sample is independent of each other and the$x_1,n_1-x_1,x_2$ and$n_2-x_2$ are all at least 5 .

Therefore, the assumptions for two proportions are satisfied.

Part(b) Step 1: Given Information

The given values are,

$$\begin{gathered} x_1=386, \\ {n}_1=750, \\ {x}_2=237, \\ {n}_2=500, \\ \alpha =0.05 \end{gathered}$$

Part(b) Step 2: Explanation 

The null hypothesis:

$H_0:p_1>p_2$

The alternative hypothesis:

$H_a:p_1>p_2$

Using the MINITAB output.

MINITAB output: Test and Cl for two proportions.

From MINITAB output, the test statistic is $1.41$and $p-$value is $0.079$

$p$- value is greater than the level of significance.

$p\text{- value}(=0.079)>\alpha (=0.05)\text{.}$

The null hypothesis is rejected.

There is no evidence to suggest that those with a bachelor's degree have a higher percentage of adults who are overweight than those with a graduate degree.

As a result, there is no evidence to suggest that those with a bachelor's degree have a higher percentage of adults who are overweight than those with a graduate degree.