Chapter 9.3, Problem 79E

(a)

To determine

To construct: and interpret a 95% confidence interval for the true hardness mean of the tablets in this batch.

Answer to Problem 79E

(11.4719, 11.5609)

There are 95 percent confident that the true hardness of the tablets in this batch is between 11.4719 and 11.5609.

Explanation of Solution

Given:

  $\begin{array}{c}n=10\\ c=95\%=0.95\end{array}$

Formula used:

Margin of error

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

Conditions

The three conditions are: Random, independent.(10% condition), Normal/Large sample.

Random: Satisfied, because the sample is a random sample.

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

Normal/ Large sample: satisfied, because the pattern in the normal quintile plot is roughly linear.

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

The mean is

  $\begin{array}{c}\overline{x}=\frac{\begin{array}{l}11.627+11.613+11.493+11.602+11.360\\ +11.374+11.592+11.458+11.552+11.463\\ +11.383+11.715+11.485+11.509+11.429\\ +11.477+11.570+11.623+11.472+11.531\end{array}}{20}\\ =\frac{230.328}{20}\\ =11.5164\end{array}$

The variance is

  $\begin{array}{c}s=\sqrt{\frac{\begin{array}{l}{\left(11.627-11.5164\right)}^2+{\left(11.613-11.5164\right)}^2+{\left(11.493-11.5164\right)}^2\\ +{\left(11.602-11.5164\right)}^2+{\left(11.360-11.5164\right)}^2+{\left(11.374-11.5164\right)}^2\\ +{\left(11.592-11.5164\right)}^2+{\left(11.458-11.5164\right)}^2+{\left(11.552-11.5164\right)}^2\\ +{\left(11.463-11.5164\right)}^2+{\left(11.383-11.5164\right)}^2+{\left(11.715-11.5164\right)}^2\\ +{\left(11.485-11.5164\right)}^2+{\left(11.509-11.5164\right)}^2+{\left(11.429-11.5164\right)}^2\\ +{\left(11.477-11.5164\right)}^2+{\left(11.570-11.5164\right)}^2+{\left(11.623-11.5164\right)}^2\\ +{\left(11.472-11.5164\right)}^2+{\left(11.531-11.5164\right)}^2\end{array}}{20-1}}\\ =0.0950\end{array}$

Determine 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=95% in the table of the Students T distribution:

  $t_{a/2}=2.093$

The margin of error is

  $\begin{array}{c}E=t_{a/2}\times \frac{s}{\sqrt{n}}\\ =2.093\times \frac{0.0950}{\sqrt{20}}\\ =0.0445\end{array}$

The boundaries of the confidence interval are

  $\begin{array}{c}\overline{x}-E=11.5164-0.0445=11.4719\\ \overline{x}+E=11.5164+0.0445=11.5609\end{array}$

There are 95% confident that the true hardness of the tablets in this batch is between 11.4719 and 11.5609.

(b)

To determine

To Explain: the interval in part (a) is consistent with the result of the test in exercise 77.

Answer to Problem 79E

Confidence interval contains 11.5

Explanation of Solution

From the Result part (a):

(11.4719, 11.5609)

It is observed that the confidence interval contains the value 11.5, which means that it is likely that the mean hardness is 11.5 and therefore it is fail to reject the claim that the mean is 11.5.

There is no enough convincing proof to help the claim that the true hardness of the tablets in this batch differs from 11.5.

We then observed that we made the same conclusion as in the previous exercise.