It it known that the duration, in hours, of a 75-watt lamp has a standard deviation of 25 hours. A random sample of 35 lamps has an average time of life of 1014 hours.
a) Build a confidence interval of 98% for the average duration of a lamp.
b) What should be the size of the sample so that the amplitude of the interval in a) was reduced to half?
My professor solved a) this way:
I don't understand why $\phi(\mathbb{z}_{0.01}) = 0.99$. In the z table I got, approximately, a value that's between -2.33 and -2.32. The average of these points is -2.325. I understand the rest but this part is bugging me. I just don't get what he did.
The professor seems to be using $z_\alpha$ to represent the upper $\alpha$ quantile of the standard normal.
$\Phi(x)$ is the cumulative distribution function (CDF) for the standard normal, i.e., $$\Phi(x) = \Pr[Z \le x]$$ for $Z \sim \operatorname{Normal}(0,1)$ and $- \infty < x < \infty.$ So when we want to find $x$ such that $\Pr[Z \le x] = 0.99$, or $\Pr[Z \le x] = 0.01$, we are computing an inverse of the CDF. Due to the symmetry of the standard normal, $$\Pr[Z \le x] = \Pr[Z \ge -x],$$ or equivalently, $$\Phi(x) + \Phi(-x) = 1.$$ So we can go about finding the critical value in multiple ways, but let us be clear on the notation. I will denote $z_{\alpha}$ to be the number that satisfies $$\Phi(z_{\alpha}) = \alpha = \Pr[Z \le z_{\alpha}].$$ That is to say, $z_\alpha$ is the $\alpha$ quantile of the standard normal. Similarly, I will denote $z^*_{\alpha}$ to be the number that satisfies $$\Phi(z^*_\alpha) = 1-\alpha = \Pr[Z > z^*_\alpha]$$ to be the upper $\alpha$ quantile of the standard normal. Then $$z^*_\alpha = -z_\alpha = z_{1-\alpha}$$ and we have for $\alpha = 0.01$ $$z_{0.01}^* = z_{0.99} \approx 2.3263478740408411009\ldots$$ If we use a table lookup, we know that $$\Pr[Z \le 2.32] \approx 0.98983, \\ \Pr[Z \le 2.33] \approx 0.990097.$$ We can get a more precise estimate of $z_{0.99}$ by using interpolation instead: we compute $$\lambda = \frac{0.99 - 0.98983}{0.990097 - 0.98983} = 0.636704,$$ hence our interpolated critical value should be $$z_{0.99} \approx (1 - \lambda)(2.32) + \lambda(2.33) = 2.32637.$$ Comparing this to the exact value shows that we get a more precise estimate to within $10^{-4}$, whereas the crude $2.325$ is actually off by $0.001$.