Solution to Exercise 7.03.

  (a) Narrower. One needs a narrower interval for less confidence. Asking for 95% confidence vs. 90%, we want to be more sure the interval contains the true value, so we have to make it wider for more confidence. This is clear from looking at the formula in the box on p. 282: we would use z_.025 = 1.960 for a 95% CI and for a 90% CI we would use the smaller multiplier z_.05 = 1.645.

(b) Not Correct. Either $\mu$ is in the interval or it is not. If we look at the r.v. version $\bar{X} \pm 1.960*\sigma/\sqrt{n}$, then there is 95% probability, but when we plug in the observed values, there is no longer any probability, so in order to express the result we use the term confidence.

(c) Not Correct. The confidence interval is a statement about the population parameter $\mu$, not about individual observations from the population. We could say that there is 95% probability that a randomly chosen bottle is within the limits $\mu \pm 1.960*\sigma$.

(d) Not Correct, probably. The number of 95% confidence intervals that contain the true $\mu$ would be a binomial r.v. with n = 100 trials and success probability p = .95. It may happen that exactly 95 trials result in ``success,'' i.e. that the true $\mu$ is in the CI, but more likely than not it would be some other value near 95. (Note: I computed P[Y=95] where $Y \sim Bin(100,.95)$, and the result was .18. So there is a .18 chance that 95 out of 100 intervals contains the true $\mu$, but a 1-.18 = .82 chance that the number of CI's containing the true value is something other than 95.)