Chapter 9.3, Problem 80E

(a)

To determine

To construct: and interpret a 90% confidence interval for the true mean vertical jump mean in inches of the student.

Answer to Problem 80E

(14.9246, 19.0754)

There are 90 percent confident that the true mean vertical jump of students at this school is between 14.9246 and 19.0754 inches.

Explanation of Solution

Given:

  $\begin{array}{l}n=20\\ c=90\%=0.90\end{array}$

Formula used:

  $E=t_{a/2}\times \frac{s}{\sqrt{n}}$

Calculation:

The three conditions for are: Random, independent and Normal/ Large sample

Random: Satisfied, because the sample of 20 students is less than 10% of the population of all students

Independent: satisfied, because the sample 20 students is less than 10 percent of the population of all students.

Normal/ Large sample: Satisfied, because the pattern in the normal quintile plot is roughly liner. This indicates that the distribution is about Normal.

Since all conditions are satisfied, it is suitable to find a confidence interval for the population mean

Confidence interval

The mean is

  $\begin{array}{c}\overline{x}=\frac{\begin{array}{l}11.0+11.5+12.5+26.5+15.0\\ +12.5+22.0+15.0+13.5+12.0\\ +23.0+19.0+15.5+21.0+12.5\\ +23.0+20.0+8.5+25.5+20.5\end{array}}{20}\\ =\frac{340}{20}\\ =17\end{array}$

The variance is

  $\begin{array}{c}s=\sqrt{\frac{\begin{array}{l}{\left(11.0-17\right)}^2+{\left(11.5-17\right)}^2+{\left(12.5-17\right)}^2+{\left(26.5-17\right)}^2+{\left(15.0-17\right)}^2\\ +{\left(12.5-17\right)}^2+{\left(22.0-17\right)}^2+{\left(15.0-17\right)}^2+{\left(13.5-17\right)}^2+{\left(12.0-17\right)}^2\\ +{\left(23.0-17\right)}^2+{\left(19.0-17\right)}^2+{\left(15.5-17\right)}^2+{\left(21.0-17\right)}^2+{\left(12.5-17\right)}^2\\ +{\left(23.0-17\right)}^2+{\left(20.0-17\right)}^2+{\left(8.5-17\right)}^2+{\left(25.5-17\right)}^2+{\left(20.5-17\right)}^2\end{array}}{20-1}}\\ =5.3680\end{array}$

Find the t-value by looking in the row starting with degrees of freedom $df=n-1=20-1=19$ and in the column with C=90% in the table of the students T distribution:

  $t_{a/2}=1.729$

The margin of error is

  $\begin{array}{c}E=t_{a/2}\times \frac{s}{\sqrt{n}}\\ =1.729\times \frac{5.3680}{\sqrt{20}}\\ =2.0754\end{array}$

The boundaries of the confidence interval

  $\begin{array}{c}\overline{x}-E=17-2.0754=14.9246\\ \overline{x}+E=17+2.0754=19.0754\end{array}$

There are 90% confident that the true mean vertical jump of students at this school is between 14.9246 inches and 19.0754 inches.

(b)

To determine

To Explain: the interval in part (a) is consistent with the result of the test in exercise 78.

Answer to Problem 80E

It is observed that we made the same conclusion as in the previous exercise.

Explanation of Solution

From the Result part (a):

(14.9246, 19.0754)

It is observed that the confidence interval contains the value 15, which means that it is likely that the mean vertical distance is 15 and therefore fails to reject the claim that the mean is 15.

There is no enough convincing evidence to support the claim that the true mean vertical distance of students at this school differs from 15 inches.

It is observed that we made the same conclusion as in the previous exercise.