Answer:
The 99% confidence interval for mean is (0.9551, 0.9569).
Step-by-step explanation:
The (1 - α)% confidence interval for the population mean is:
[tex]CI=\bar x\pm z_{\alpha/2}\ \frac{\sigma}{\sqrt{n}}[/tex]
The information provided is:
[tex]n=33\\\bar x=\mu=0.956\\\sigma=0.002\\(1-\alpha) \%=99\%[/tex]
The law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample ([tex]\bar x[/tex]) approaches the whole population mean (µ).
In this case the sample size is quite large, i.e. n = 33 > 30. So, according to the law of large numbers, the sample mean is approximately same as the population mean.
The critical value of z for 99% confidence level is:
[tex]z_{\alpha/2}=z_{0.005}=2.58[/tex]
*Use a z-table.
Compute the 99% confidence interval for mean as follows:
[tex]CI=\bar x\pm z_{\alpha/2}\ \frac{\sigma}{\sqrt{n}}[/tex]
[tex]=0.956\pm 2.58\times \frac{0.002}{\sqrt{33}}\\=0.956\pm 0.0009\\=(0.9551, 0.9569)[/tex]
Thus, the 99% confidence interval for mean is (0.9551, 0.9569).
