Question

Suppose the investigator believes that virtually all values of breakdown voltage are between 40 and 70. What sample size would be appropriate for the 95% CI to have a width of 2 kV? (so that m is estimated to within 1 kV with 95% confidence)

Answer 1

Answer:

The sample size required is 865.

Step-by-step explanation:

The (1 - α)% confidence interval for population mean is:

$CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

The margin of error of this interval is:

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

It is provided that all values of breakdown voltage are between 40 and 70.

So, the minimum value of breakdown voltage is, Min. = 40 and the maximum value of breakdown voltage is, Max. = 70.

Assume that the population standard of the distribution of breakdown voltage is known.

The standard deviation is:

$\sigma=\frac{Range}{4}=\frac{Max.-Min.}{4}=\frac{70-40}{2}=15$

The margin of error is, MOE = 1 kV.

The critical value of z for 95% confidence level is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the sample size as follows:

$MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}\\1=1.96\times \frac{15}{\sqrt{n}}\\n=(\frac{1.96\times 15}{1})^{2}\\n=864.36\\n\approx865$

Thus, the sample size required is 865.