Respuesta :
Answer:
The sample size required is 22.
Step-by-step explanation:
The random variable X can be defined as the scale readings of a lab scale.
The random variable X is normally distributed with standard deviation, σ = 0.0002 grams.
The procedure is repeated five times, i.e. the sample size is, n = 5.
The sample mean reading is, [tex]\bar x=10.0023[/tex].
The (1 - α)% confidence interval for the population mean is:
[tex]CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}[/tex]
The margin of error of this interval is:
[tex]MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}[/tex]
The critical value of z for 98% confidence interval is:
[tex]z_{\alpha/2}=z_{0.02/2}=z_{0.01}=2.33[/tex]
*Use a z-table.
Compute the sample size required to get a margin of error of 0.0001 as follows:
[tex]MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}\\0.0001=2.33\times \frac{0.0002}{\sqrt{n}}\\n=(\frac{2.33\times 0.0002}{0.0001})^{2}\\n=21.7156\\n\approx22[/tex]
Thus, the sample size required is 22.
