Chapter 23, Problem 18E

(a)

To determine

To calculate what is the standard error of the mean for these data.

Answer to Problem 18E

The standard error is $7.90$ .

Explanation of Solution

It is given that after his first attempt to determine the speed of light, Michelson conducted an improved experiment. Also, it is given that,

  $\begin{array}{l}\mu =852.4\\ \sigma =79.0\\ n=100\end{array}$

As we know that the standard error of the mean is the standard deviation divided by the square root of the sample size. Thus, we have that,

  $\begin{array}{l}SE=\frac{\sigma }{\sqrt{n}}\\ =\frac{79.0}{\sqrt{100}}\\ =7.90\end{array}$

Therefore, the standard error is $7.90$ .

(b)

To determine

To explain how would you expect a $95\%$ confidence interval for the second experiment to differ from the confidence interval for the first, without computing it.

Explanation of Solution

It is given that after his first attempt to determine the speed of light, Michelson conducted an improved experiment. Also, it is given that,

  $\begin{array}{l}\mu =852.4\\ \sigma =79.0\\ n=100\end{array}$

Thus, the confidence interval will be centered about a higher value because the mean of $852.4$ is larger than the mean of $756.22$ . The confidence interval will be narrower because both the standard deviation is smaller and the sample size is larger for this confidence interval than for the confidence interval of previous exercise.

The confidence interval will be narrower because the standard error found in previous exercise is $7.90$ , which is smaller than the standard error of the exercise $21$ .

(c)

To determine

To explain using your confidence interval that what does this indicate about Michelson’s two experiments.

Answer to Problem 18E

The experiment of $1882$ appears to be more valid than the experiment of $1897$ .

Explanation of Solution

It is given that after his first attempt to determine the speed of light, Michelson conducted an improved experiment. Also, it is given that,

  $\begin{array}{l}\mu =852.4\\ \sigma =79.0\\ n=100\end{array}$

Thus, experiment of exercise $21$ : The confidence interval of exercise $21$ contains the value of $710.5$ , which indicates that the corresponding experiment could measure the speed of light in a correct manner.

Experiment of exercise $22$ : The value of $710.5$ is more than two standard deviations ( $7.90$ ) of the mean ( $852.4$ ), which indicates that the corresponding experiment appears to not measure the speed of light in a correct manner (because it does not contain the actual speed of light).

Conclusion: The experiment of $1882$ appears to be more valid than the experiment of $1897$ .