Hypothesis Test-Confidence Interval connection

Transcription

1 Hyothesis Test-Confidence Interval connection Hyothesis tests for mean Tell whether observed data are consistent with μ = μ. More secifically An hyothesis test with significance level α will reject the hyothesis μ = μ if and only if P(observed data μ = μ ) < α. Confidence Intervals for mean Give a set of values for true mean, μ, that are not inconsistent with the observed data. More secifically If μ is a value outside the (1- α)% confidence interval, then P(observed data μ = μ ) < α. From the above we can infer If μ is a value outside the (1- α)% confidence interval, then an hyothesis test with significance level α will reject the null hyothesis H : μ = μ. Similarly, if μ is inside the confidence interval then H : μ = μ will not be rejected.

2 Examle 1: Facial Height Data The average change in facial height in the samle was -1.7 mm, with s = 6.8 mm. The null hyothesis, H : μ =, was rejected at the α =.5 level. The 95% confidence interval for the mean change in facial height is ( 3.2,.2), which does not contain. Examle 2: Chewing Gum Data The average two-year change in DMFS for those in grou A was.72 DMFS. The 95% confidence interval was (-2.92, 1.48), which contains. The null hyothesis H : μ = was not rejected ( =.51). Note that the 95% confidence interval shows that neither a true average decrease of more than 2 DMFS nor an increase of more than 1 DMFS would be inconsistent with these data. Thus, not rejecting a null hyothesis does not necessarily indicate that we should accet the null hyothesis as the truth!

3 Hyothesis Test for Proortions: The one-samle test of roortions is based on the statistic Z ˆ (1 ) / n. If Z is large (Z is far from zero), it suggests that. If = and n(1-) > 5, then Z will have an aroximate N(,1) distribution. If Z is larger value than one would exect to come from a N(,1) distribution, then we will have evidence that.

4 Examle: Presidential Election Poll Presidential reference oll results (7/5/216)* Trum Clinton Total *among those who favored either Trum or Clinton Do these data rovide evidence that Trum has a lead in the oulation samled? To answer this question we can comute the hyothesis test H : <.5, and H 1 : >.5, where is the roortion referring Trum. Note that the form of H differs from reviously introduced tests. With these hyotheses, a rejection of H will indicate evidence that Trum has a lead If we used H : =.5, H 1 :.5, a rejection of H would indicate only that the candidates are not tied. This is an examle of a one-sided hyothesis test.

5 Examle: Presidential Election oll continued To test the one-sided hyotheses H : <.5, and H 1 : >.5, Comute the Z statistic: ˆ 4/ Z ˆ (1 )/ n / Note that for we use the border robability,.5. This would be the worst case in terms of disroving H. For one-sided tests the -value is also one-sided. For this examle the -value is: -value = P( N(,1) >.726 ) =.234 The robability is on only one side of the distribution. The side it is on should be the side that indicates that H is not true.

6 Examle: Presidential Election oll continued For this examle, let α =.5, then since the -value =.234 is greater than.5 we do not reject H. Suose we cannot comute the -value easily we can also comare the statistic to a critical value to see if the -value is less than or greater than α. In the case of a one sided test at significance level α: o If H 1 : >, then reject H if Z > Z 1-α o If H 1 : <, then reject H if Z < Z α Note that we do not halve the α for one-sided tests. So for this examle we would comare Z =.726 to Z.95 = Since.726 < 1.645, we do not reject H.

7 Final notes on the Presidential Poll Examle: Though we did not reject H, this does NOT necessarily indicate that H 1 is true. The 95% confidence interval for is (47.8%, 54.8%), which indicates that there is NOT strong evidence of either candidate having the majority. Tests for roortions do not have to be one-sided. Two-sided tests for roortions can also be formulated. One-sided tests can be used for comarisons other than roortions (e.g. one can use them with t-tests).

8 One-Samle statistics summary Binary data (a.k.a. dichotomous, or yes/no data) o Large samle, n(1-) > 5 Confidence interval for roortion, 8.3 One-samle test for roortion, 9.3 o Small samle, n(1-) < 5 Exact Binomial confidence intervals, HW #6 Exact one-samle Binomial Test* Continuous data (a.k.a. Interval data) o Large samle Confidence interval for mean 8.2 One-samle t test 9.2 o Small samle, data from Normal distribution Confidence interval for mean 8.2 One-samle t test 9.2 non-normal data *not covered in course toics Exact confidence interval for median* One-samle test for median (similar to sign test 13.1

9 One-samle T-test 1. Decide on hyotheses: Two sided: H : μ = μ, vs. H 1 : μ μ One sided: H : μ μ, vs. H 1 : μ > μ, or H : μ μ, vs. H 1 : μ < μ 2. Decide on significance level α 3. Comute the statistic X T s n 4. Rejection criteria Two sided test: reject H if T > t n-1,1-α/2 One sided test: If H 1 : μ > μ, reject H if T > t n-1,1-α If H 1 : μ < μ, reject H if T < -t n-1,1-α 5. Comute -value Two sided test: -value = P( t n-1 > T ) One sided test: If H 1 : μ > μ, then -value = P(t n-1 > T) If H 1 : μ < μ, then -value = P(t n-1 < T) 6. Confidence interval for μ X t s n X t s n n1,1 2, n1,1 2

10 One-samle Z-test for roortions Only valid if n(1-) > 5. Estimate using ˆ. 1. Decide on hyotheses: Two sided: H : =, vs. H 1 : One sided: H :, vs. H 1 : >, or H :, vs. H 1 : < 2. Decide on significance level α 3. Comute the statistic Z ˆ (1 ) / n 4. Rejection criteria Two sided test: reject H if Z > Z 1-α/2 One sided test: If H 1 : >, then reject H if Z > Z 1-α If H 1 : <, then reject H if Z < -Z 1-α 5. Comute -value Two sided test: -value = P( N(,1) > Z ) One sided test: If H 1 : >, then -value = P(N(,1) > Z) If H 1 : <, then -value = P(N(,1) < Z) 6. Confidence interval for : Z ˆ(1 ˆ) n, ˆ Z ˆ(1 ˆ ) ˆ 1 / 2 1 / 2 n

11 Using Table 4 to estimate t robabilities Because Table 4 in the courseack only gives ercentiles of selected t distributions, we cannot get exact t robabilities from them. We can get bounds on the robabilities. Examle: Use table 4 to estimate P(t 83 > 2.29). Table 4 imlies that 1.99 and 2.37, are the 97.5 th and 99 th ercentiles of the t 83 distribution. Which says: o P(t 83 > 1.99) =.25, and o P(t 83 > 2.37) =.1. Since 2.29 is between the 1.99 and 2.37, it follows that P(t 83 > 2.29) is between P(t 83 > 1.99) and P(t 83 > 2.37). P(t 83 > 1.99) = > P(t 83 > 2.29) >.1 P(t 83 > 2.37) = So P(t 83 > 2.29) is between.1 and.25.