Chapter 21, Problem 13E

(a)

To determine

To explain we cannot use this information to construct a confidence interval for the rate of occurrence of early hip dysplasia among all six-month old puppies.

Explanation of Solution

It is given in the question that canine hip dysplasia is a degenerative disease that causes pain in many dogs. A veterinarian checked $42$ puppies whose owners brought them to a vaccination clinic and she found five with early hip dysplasia. Thus, let us check the appropriate conditions and assumptions for inference to proceed further to test hypothesis as:

Random condition: It is satisfied as we assumed that the sample was randomly selected from the population.

Independent condition: It is satisfied as we can assume that the sample is independent.

  $10\%$ condition: It is satisfied as the sample is less than $10\%$ of all population.

Success/failure condition: It is not satisfied because both are not greater than ten as,

  $\begin{array}{l}np=42(\frac{5}{42})\\ =5<10\\ n(1-p)=42(1-\frac{5}{42})\\ =37>10\end{array}$

Thus, all the conditions are not satisfied. So, we cannot use this information to construct a confidence interval for the rate of occurrence of early hip dysplasia among all six-month old puppies.

(b)

To determine

To construct a “plus-four” confidence interval and interpret it in this context.

Answer to Problem 13E

We are $95\%$ confident that between $5\%\text{ and 26\%}$ of puppies will show signs will show signs of hip dysplasia.

Explanation of Solution

It is given in the question that canine hip dysplasia is a degenerative disease that causes pain in many dogs. A veterinarian checked $42$ puppies whose owners brought them to a vaccination clinic and she found five with early hip dysplasia. And in the Success/failure condition: It is not satisfied because both are not greater than ten as,

  $\begin{array}{l}np=42(\frac{5}{42})\\ =5<10\\ n(1-p)=42(1-\frac{5}{42})\\ =37>10\end{array}$

Thus, all the conditions are not satisfied. So, we cannot use this information to construct a confidence interval for the rate of occurrence of early hip dysplasia among all six-month old puppies. So, let us construct a “plus-four” confidence interval as:

Let us add two to success and failure condition as,

  $\begin{array}{l}\frac{5+2}{42+4}=\frac{7}{46}\\ =0.1522\end{array}$

And now, to find the confidence interval we will use the calculator $\text{TI}-89$ . So, we have to first go to the $\text{Ints}$ menu and select $5:1-\text{PropZInt}$ . Then we have to enter the number of successes observed and the sample size. Then specify a confidence level. Then the interval will be calculated.

  $\begin{array}{l}x=7\\ n=46\\ c=0.95=95\%\end{array}$

Thus, the confidence interval will be as:

  $(0.0483,0.2561)$ .

Thus, we are $95\%$ confident that between $5\%\text{ and 26\%}$ of puppies will show signs will show signs of hip dysplasia.