Confidence Interval Problems
MA 223
√
1. Use the fact that Z = (x̄ − µ)/(σ/ n) will be approximately standard normal, regardless of the underlying distribution of the widget diameter, since n ≥ 30. Also, me may
use σ = s here, again since the sample size is large. Since −1.96 < Z < 1.96 with 95
percent probability, a little algebra leads to confidence interval
√
√
5.3 − (1.96)(0.12/ 100) < µ < 5.3 + (1.96)(0.12/ 100)
or 5.276 < µ < 5.323.
2. Same as the last problem but use P (Z > −1.645) = 0.95, leading to (5.3 − µ)/0.012 >
−1.645 with “95 percent confidence,” or
µ < 5.32
after algebra. Note that if you start with Z < 1.645 you’ll end up with a confidence
interval of the form µ > M for some M , not what you want.
3. We can use the normal approximation since np̂ > 10, n(1 − p̂) > 10, where p̂ is the
sample proportion.
√
The random variable Z = (p̂ − p)/ p(1 − p)/1000 is approximately standard normal.
Start with Z > −1.645 (95 percent confidence), which leads to
√
(0.028 − p)/ p(1 − p)/1000 > −1.645.
Use Maple to find that value of p that makes the inequality above true. You find
p < 0.038. By the way, if you start with Z < 1.645 you end up with a confidence
interval of the form p > M for some M ,√
not what you want. You may find it helpful
to plot the quantity Z = (0.028 − p)/ p(1 − p)/1000 in Maple to easily see that
Z > −1.645 does in fact correspond to an interval of the form 0 < p < M .
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