Chapter 14.5, Problem 30E

To determine

To find the size of the sample used to determine the given values.

Answer to Problem 30E

  $100$

Explanation of Solution

Given:

There is a probability of $0.99$ that the average life of a disposable hand warming package is between $9.7936$ and $10.2064$ hours. The standard deviation of the sample is $0.8$ hours.

Calculation:

It is known thatthe range of the interval about the mean that has probability for the sample having parameters $s,N$ and $\overline{X}$ is $\left[\overline{X}-t{\sigma }_{\overline{X}},\overline{X}+t{\sigma }_{\overline{X}}\right]$

Here the value of $t$ (number of standard deviations) is corresponding value of the probability $P$

Because the probability is $0.99$ , corresponding value of is $2.58$

By comparing the intervals we get

  $\left[\overline{X}-t{\sigma }_{\overline{X}},\overline{X}+t{\sigma }_{\overline{X}}\right]=[9.7936,10.2064]$

Thus

  $\begin{array}{l}\overline{X}-(2.58){\sigma }_{\overline{X}}=9.7936\\ \overline{X}+(2.58){\sigma }_{\overline{X}}=10.2064\end{array}$

From the above two equations, we can write by subtracting one from the other as

  ${\sigma }_{\overline{X}}=0.08$

We know that, if a sample of data has values $N$ and $s$ is the standard deviation, then the standard error of the mean is

  $s_{\overline{X}}=\frac{s}{\sqrt{N}}$ ...... (1)

Substitute the values $s=0.8,s_{\overline{X}}=0.08$ in (1) we get

  $\begin{array}{l}{s}_{\overline{X}}=\frac{s}{\sqrt{N}}\\ 0.08=\frac{0.8}{\sqrt{N}}\\ N=100\end{array}$

Therefore the size of the sample used to determine the above values is $100$ .