Econ 379: Business and Economics Statistics Instructor: Yogesh Ual Email: yual@ysu.edu
Chater 9, Part A: Hyothesis Tests Develoing Null and Alternative Hyotheses Tye I and Tye II Errors Poulation Mean: σ Known Poulation Mean: σ Unknown
Develoing Null and Alternative Hyotheses Hyothesis testing can be used to determine whether a statement about the value of a oulation arameter should or should not be rejected. The null hyothesis, denoted by H, is a tentative assumtion about a oulation arameter. The alternative hyothesis,, denoted by H a, is the oosite of what is stated in the null hyothesis. The alternative hyothesis is what the test is attemting to establish.
Summary of Forms for Null and Alternative Hyotheses about a Poulation Mean The equality art of the hyotheses always aears in the null hyothesis. In general, a hyothesis test about the value of a oulation mean μ must take one of the following three forms (where μ is the hyothesized value of the oulation mean). H H a : μ = : μ < μ μ H H a : μ = μ : μ > μ H H a : μ = μ : μ μ One-tailed (lower-tail) One-tailed (uer-tail) Two-tailed
Examle: Air Quality Suose I am interested in checking if the average PMI in the oulation is 5. Null Hyothesis H : μ = Alternative Hyothesis H H H : μ < 5 : μ > 5 5 μ = the average air quality in the oulation a a a : μ 5 One-tailed (lower-tail) One-tailed (Uer-tail) Two-tailed
Tye I Error A Tye I error is rejecting H when it is true. Tye II Error A Tye II error is acceting H when it is false. Statisticians avoid the risk of making a Tye II error by using do not reject H and not accet H.
Tye I and Tye II Errors Poulation Condition Conclusion Accet H (Conclude μ < 12) H True (μ < 12) Correct Decision H False (μ > 12) Tye II Error Reject H (Conclude μ > 12) Tye I Error Correct Decision
Some Definitions: Level of Significance: The robability of making Tye I error. Critical Value: The value (determined by the level of significance) that establishes the boundary of the rejection region. Test Statistic: A comuted value which is comared to the critical value to reject or not reject the null. -value: is the robability of getting a value more extreme than the test statistic.
One Samle z-test: z Stes of Hyothesis Testing When σ is known Ste 1. Develo the null and alternative hyotheses. Ste 2. Secify the level of significance α.. This will define the critical value for the test. Ste 3. Comute the value of the test statistic (z) or the -value corresonding to that test statistic.
Stes of Hyothesis Testing When σ is known Ste 4. Lower Tailed test ( H a : μ < μ) : Reject H if z z α or -value. Uer Tailed test : μ > μ ) : z +z α ( H a α α Reject H if or -value. Two-Tailed Tailed test ( H a : μ μ) : Reject H if z z α / 2 or z +z α / 2 or -value. α
Hyothesis Testing When σ is known The test statistic in this case is given by x μ x μ z = = σ σ / x n
Lower-Tailed Test About a Poulation Mean: σ Known -Value < α, so reject H. α =.5 -value =.82 z z z α, so reject H. z = -2.4 -z α = -1.65
Uer-Tailed Test About a Poulation Mean: σ Known -Value < α, so reject H. α =.5 -Value =.11 z α = 1.65 z = 2.29 z z z α, so reject H.
Two-Tailed Tailed Tests About a Poulation Mean: σ Known 1/2 -value =.31 1/2 -value =.31 α/2 =.25 z = -2.74 -z α/2 = -1.96 α/2 =.25 z z = 2.74 z α/2 = 1.96
Examle of Lower-Tailed Test: Air Quality Data Suose you are interested in checking if the air quality is good. Suose the level of significance ( ) is 5%. Suose a samle of size 5 is selected. Ste 1: α H : μ 5 H a : μ < 5 Ste 2: The critical value corresonding to =.5 is -1.65 α
Examle of Lower-Tailed Test: Air Quality Data Ste 3: Comute the value of test statistic. Ste 4:?????? x μ z = σ / n
Tests About a Poulation Mean: σ Unknown Test Statistic t = x μ s / n This test statistic has a t distribution with n - 1 degrees of freedom.
Tests About a Poulation Mean: σ Unknown Rejection Rule H a : μ < μ H a : μ > μ H a : μ μ Reject H if t < -t α or value < α Reject H if t > t α or value < α Reject H if t < - t α/2 or t > t α/2 or value < α
Examle: Highway Patrol One-Tailed Test About a Poulation Mean: σ Unknown At Location F, a samle of 64 vehicles shows a mean seed of 66.2 mh with a standard deviation of 4.2 mh. Use α =.5 to test the hyothesis.
One-Tailed Test About a Poulation Mean: σ Unknown 1. Determine the hyotheses. H : μ= = 65 H a : μ > 65 2. Secify the level of significance. α =.5 3. Comute the value of the test statistic. t x μ 66.2 65 = = = s / n 4.2/ 64 2.286
One-Tailed Test About a Poulation Mean: σ Unknown 4. Determine the critical value and rejection rule. For α =.5 and d.f. = 64 1 = 63, t.5 = 1.669 Reject H if t > 1.669 5. Determine whether to reject H. Because 2.286 > 1.669, we reject H. We are at least 95% confident that the mean seed of vehicles at Location F is greater than 65 mh. Location F is a good candidate for a radar tra.
A Summary of Forms for Null and Alternative Hyotheses About a Poulation Proortion The equality art of the hyotheses always aears in the null hyothesis. In general, a hyothesis test about the value of a oulation roortion must take one of the following three forms (where is the hyothesized value of the oulation roortion). H : = H a : < H : = H a : > H H a : : = One-tailed (lower tail) One-tailed (uer tail) Two-tailed
Tests About a Poulation Proortion Test Statistic where: z = σ ( 1 ) σ = n assuming n > 5 and n(1 ) > 5
Tests About a Poulation Proortion Rejection Rule H a : > H a : < H a : Reject H if z > z α or value < α Reject H if z < -z α or value < α Reject H if z < -z α/2 or z > z α/2 or value < α
Two-Tailed Tailed Test About a Poulation Proortion Examle: National Safety Council For a Christmas and New Year s s week, the National Safety Council estimated that 5 eole would be killed and 25, injured on the nation s s roads. The NSC claimed that 5% of the accidents would be caused by drunk driving.
Two-Tailed Tailed Test About a Poulation Proortion Examle: National Safety Council A samle of 12 accidents showed that 67 were caused by drunk driving. Use these data to test the NSC s claim with α =.5.
Two-Tailed Tailed Test About a Poulation Proortion 1. Determine the hyotheses. H : =.5 H : a.5 2. Secify the level of significance. α =.5 3. Comute the value of the test statistic. a common error is using in this formula σ (1 ).5(1.5) = = =.45644 n 12 (67 /12).5 z = = = σ.45644 1.28
Value Aroach Two-Tailed Tailed Test About a Poulation Proortion 4. Comute the -value. For z = 1.28, cumulative robability =.8997 value = 2(1.8997) =.26 5. Determine whether to reject H. Because value =.26 > α =.5, we cannot reject H.
Two-Tailed Tailed Test About a Poulation Proortion Critical Value Aroach 4. Determine the critical value and rejection rule. For α/2 =.5/2 =.25, z.25 = 1.96 Reject H if z < -1.96 or z > 1.96 5. Determine whether to reject H. Because -1.96< 1.278 < 1.96, we cannot reject H.