Q. T1.12 A study among the Pima Indians o... [FREE SOLUTION]

91影视

The Practice of Statistics for AP Examination

Daren Starnes, Josh Tabor

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 837 pages

ISBN: 9781319113339

Chapter 1: Q. T1.12 (page 1)

A study among the Pima Indians of Arizona investigated the relationship between a mother鈥檚 diabetic status and the number of birth defects in her children. The results appear in the two-way table.
a. What proportion of the women in this study had a child with one or more birth defects?
b. What percent of the women in this study were diabetic or prediabetic, and had a child with one or more birth defects?
c. Make a segmented bar graph to display the distribution of number of birth defects for the women with each of the three diabetic statuses.
d. Describe the nature of the association between mother鈥檚 diabetic status and number of birth defects for the women in this study.

Short Answer

Expert verified

Part a. $0.0439$

Part b. $1.823 %$

Part c. The width of the every bar has to be the same and the height has to be equal to the percent.

Part d. The number of birth defects gets to be highest for the Diabetics and lowest for the Non diabetics.

Step by step solution

Part a. Step 1. Given information

Given:

Part a. Step 2. Calculation

Suppose first find the row/ column totals, which is the sum of all counts in the associating row/ column.

It is observed the table is having the data about $1207$ women and that $53$ of the $1207$ women had a child with one or more birth defects (as $53$ is mentioned in the row 鈥淥ne or more鈥� and in the column 鈥淩ow Total鈥� of the above table).

$\frac{53}{1207} = 0.0439$

Therefore is a proportion of $0.0439$ women had a child with one or more birth defects.

Part b. Step 1. Calculation

Let us first determine the row/column totals of the given table, which is the sum of all counts in the associating row/ column.

It is observed that the table is having the data about $1207$ women and that $13$ of the $1207$ women were pre diabetic and had a child with one or more birth defects where $9$ of the $1207$ women were diabetic and had a child with one or more birth defects

$\frac{13 + 9}{1207} = \frac{22}{1207} = \frac{0.01823 = 1.823 %}{}$

Therefore $1.823 %$ of the women in the study was pre diabetic, and has a child with one or more birth defects.

Part c. Step 1. Calculation

Let us first find the row/column totals of the given table, which is the sum of all counts in the associating row/ column.

To find the distribution of superpower preference from every country, and divide the counts by the column total.

Graph:

The width of the every bar has to be the same and the height has to be equal to the percent.

Part d. Step 1. Calculation

From the part (c)

It is observed that the number of birth defects gets to be highest for the Diabetics and lowest for the Non diabetics, the reason is that the green area of the bars is greatest for Diabetic and least for non diabetic.