Chapter 1, Problem T1.12SPT

(a)

To determine

To find: the proportion of the women in this study had a child with one or more birth defect.

Answer to Problem T1.12SPT

0.0439

Explanation of Solution

Given:

	Non diabetic	Pre diabetic	Diabetic
Non	754	362	38
one or more	31	13	9

Calculation:

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

	Non diabetic	Pre diabetic	Diabetic	total
Non	754	362	38	1154
one or more	31	13	9	53
total	785	375	47	1207

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 “One or more” and in the column “Row Total” 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.

(b)

To determine

To find: the percent of the women in this study were diabetic or pre diabetic.

Answer to Problem T1.12SPT

1.823%

Explanation of Solution

Given:

	Non diabetic	Pre diabetic	Diabetic
Non	754	362	38
one or more	31	13	9

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.

	Non diabetic	Pre diabetic	Diabetic	total
Non	754	362	38	1154
one or more	31	13	9	53
total	785	375	47	1207

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

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

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

(c)

To determine

To construct: a segmented bar graph to show the distribution of number of birth defects for the women with every three diabetic statues.

Explanation of Solution

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.

	Non diabetic	Pre diabetic	Diabetic	total
Non	754	362	38	1154
one or more	31	13	9	53
total	785	375	47	1207

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

	Non diabetic	Pre diabetic	Diabetic
None	96.05%	96.53%	80.85%
One or more	3.95%	3.47%	19.15%

Graph:

  

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

(d)

To determine

To Explain: the nature of the association between mother’s diabetic status and number of birth defects for the women in this study.

Answer to Problem T1.12SPT

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

Explanation of Solution

Given:

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.