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February 22, 2006, Chapter 16, Inference about mean,
unknown standard deviation
HOMEWORK Due Feb. 24: 13.25-13.27, 14.7-14.8. 14.11, 14.12, 14.14, 14.15, 14.21, 14.26, 14.27, 14.29 plus excel assignment 3. Due Feb. 24: 14: 34, 35, 37. 15: 18- Due March 3rd: 15: 26, 27, 29. Midterm II: March 3rd.
NOTE on calculating P -value. Let z¯ be the z-statistic for your example. Three cases:
- If Ha : μ > c then P -value is P (z ≥ z¯).
- If Ha : μ < c then P -value is P (z ≤ z¯).
- If Ha : μ 6 = c then P -value is P (z ≥ |z¯|) + P (z ≤ −|z¯|)
Review Terminology about hypothesis testing
- Power of a test: Fix α significance level. The power of a test is the probability of rejecting H 0 when μ has some particular alternative value.
- Type I error: reject H 0 when H 0 is actually true.
- Type II error: fail to reject H 0 when H 0 is false.
Example: (15.32) If μ = 134, probability of rejecting H 0?
- Reject H 0 if |z| ≥ 1. 96. (from α =. 05 )
- True if either x¯ ≥ 131. 46 or x¯ ≤ 124. 535.
- Normalize using μ = 134 : 13115.^46 /√− 72134 = − 1. 436. P (¯x ≥ 131 .46) = P (z ≥ − 1 .436) =. 924.
- 12415.^535 /√− 72134 = − 5. 35. P (¯x ≤ 124 .535) = P (z ≤ − 5 .35) ≤. 00001.
So power is. 924.
Type I error: probability. 05
Type II error: probability. 076.
Recall: σ known, then the z-statistic
z =
x¯ − μ σ/
n
is N (0, 1).
When σ is unknown, we substitute the sample standard deviation, s, for the population standard deviation, σ.
The t-statistic is
t =
x¯ − μ s/
n
As we run through all samples of size n, t has the t distribution with n − 1 - degrees of freedom.
These curves are shaped similarly to the normal curve, but not exactly the same. As the number of degrees of freedom goes up, the curves get closer to the standard normal curve.
Cannot use Table A for t-distribution. Can use Table C, or software (or TI-83).
Remember that t = (^) s/x¯−√μ 19 has the t-distribution with 18 degrees of freedom.
So 95% of the time, t is between − 2. 101 and 2. 101. So for 95% of samples,
x¯ − μ s/
Solving we get
μ − 2. 101 · s ·
19 ≤ x¯ ≤ μ + 2. 101 · s ·
Equivalently, μ is in x¯ ± 2. 101 · s/
19 with 95% confidence.
So our 95% confidence interval is
- 811 ±. 0824.
Example: Estimate mean monthly cell phone bills in 2004. Give a 90% confidence interval.
We have a sample of 12 monthly phone bills: 94.25 38.94 79.15 56.78 70.07 115.59 77.56 37.01 55.00 76.05 27.29 52.48.
x ¯ = 65. 01 , s = 25. 46
Use t-distribution with 11 -degrees of freedom. To get critical value for 90%, look up. 05 in Table C on line 11. t∗^ = 1. 796.
μ is $65. 01 ± 1. 796 · 25. 46 /
Example: In NEJM study, sample of about 16,000 women aged 50-79 asked to take calcium supplements over 7 year period. 16 hip fractures in this group.
In population, would expect 22.4 hip fractures in this size group over same period. Is calcium supplementation effective in reducing chances of hip fracture at α. 05?
The mean rate of hip fractures is .0014 in the general population. (14 per 10, over 7 years in 50-79 year old women).
Let μ be the mean rate of hip fractures among women if they take calcium supplements.
- H 0 : μ =. 0014. The null hypothesis is that calcium treatment has no effect. Ha : μ <. 0014. The alternative hypothesis is that calcium treatment reduces the risk of hip fractures.
- Calculate the test statistic. x¯ =. 001 , s =. 032
t =
x¯ −. 0014
. 0316 /
- P-value: At 15,999 df, the t-statistic is indistinguishable from the z-statistic (see Table C and how close the t-statistic and z-statistic are already at 1000 df).
So from table C, the P-value is between .05 and .1 (from Table A it is .0548).
- Conclusion: This test is not statistically significant, we cannot conclude that calcium supplments reduce hip fractures. reject H 0
