Chapter 10, Problem 10.68RE

(a)

To determine

To Explain: that it would be reasonable to construct a confidence interval based on z-critical values here, in spite of the histogram’s shape.

Answer to Problem 10.68RE

Sample size is sufficiently large.

Explanation of Solution

Given:

  $\begin{array}{l}c=95\%=0.95\\ n=164\\ \overline{x}=0.285\\ s=0.3\end{array}$

The sample size of $n=164$ is large and the centre limit theorem then tells that the sampling distribution of the sample mean is approximately normal. This then implies that it can construct a confidence interval using the $z$ -critical values.

(b)

To determine

To Calculate: the typical mercury concentration in cans of tuna sold in stores and construct and interpret a 95% confidence interval.

Answer to Problem 10.68RE

(0.2391, 0.3309)

There is 95% confidence that the mean mercury content in cans of tuna sold in store is between $0.2391$ ppm and $0.3309$ ppm.

Explanation of Solution

Formula used:

  $E=z_{\alpha /2}\times \frac{\sigma }{\sqrt{n}}$

For confidence intervals are

  $\begin{array}{l}\overline{x}-E\\ \overline{x}+E\end{array}$

Calculation:

For confidence level $1-\alpha =0.95$ , $z_{\alpha /2}=z_{0.025}$ using the normal probability table

  $z_{\alpha /2}=1.96$

The margin of error is

  $E=z_{\alpha /2}\times \frac{\sigma }{\sqrt{n}}=1.96\times \frac{0.3}{\sqrt{164}}=0.0459$

Confidence intervals are

  $\begin{array}{l}\overline{x}-E=0.285-0.0459=0.2391\\ \overline{x}+E=0.285+0.0459=0.3309\end{array}$

There is 95% confidence that the mean mercury content in cans of tuna sold in store is between $0.2391$ ppm and $0.3309$ ppm