HYPOTHESIS TESTS FOR ONE POPULATION MEAN WORKSHEET MTH 1210, FALL 2018 We are resposible for 2 types of hypothesis tests that produce ifereces about the ukow populatio mea, µ, each of which has 3 possible alterative hypothesis (6 cases total). As is the case with cofidece itervals, our ifereces are made with a prescribed certaity called the cofidece level. For cofidece itervals, we describe this with our cofidece level (e.g. 90%, 95% or 99%). Whe we do hypothesis tests, we idicate our probability of makig a error, which is α. The probability α is called our sigificace level. It is crucial that α is determied prior to the hypothesis testig procedure; i fact, we should really choose α before observig the data i the sample i order for our uderlyig assumptios to be satisfied. The sigificace level α should be stated at the start of ay hypothesis test. For ay hypothesis testig procedure we the state a ull hypothesis, H 0. For the hypothesis testig procedure o this worksheet, H 0 is the hypothesis that the populatio mea, µ, is equal to a umber. We ofte thik about this umber, deoted µ 0, as the presumed populatio mea. We use the hypothesis test to determie if there is strog evidece that this ull hypothesis is icorrect. The ext step of the hypothesis test is our alterative hypothesis, H a. This is a hypothesis that is cotrary to our ull hypothesis; if there is eough evidece for the alterative hypothesis, the the ull hypothesis is rejected. The coclusio of our hypothesis tests are always either reject the ull hypothesis or fail to reject the ull hypothesis. As usual, the size of our sample is, mea x ad stadard deviatio is deoted s. Each of our 6 hypothesis tests uses a test statistic computed i terms of these sample statistics (ad, possibly, the populatio stadard deviatio σ). We coclude our hypothesis test usig oe of the followig two approaches: Optio I, Critical Value Approach: Usig this approach, we idetify a rejectio regio that is based o the z or t critical values from Table IV i our textbook Appedix A. If the test statistic is i that rejectio regio, we reject the ull hypothesis ; otherwise, we fail to reject the ull hypothesis. The umber α represets the (small) probability that we reject the ull hypothesis eve though it is actually true. Optio II, P -value Approach: Usig this approach, we idetify a P -value that is calculated from our test z or t test statistics usig Table II or from the Detailed t-table Areas to the right of t, table passed out i class, respectively. If the P value is less tha the umber α, we reject the ull hypothesis ; otherwise, we fail to reject the ull hypothesis. The umber α represets the (small) probability that we reject the ull hypothesis eve though it is actually true. 1. z-tests for the populatio mea (σ is kow) For these tests, we assume that σ is kow. We also assume that is relatively large (> 30) or that the uderlyig populatio is roughly ormal. 1
2HYPOTHESIS TESTS FOR ONE POPULATION MEAN WORKSHEET MTH 1210, FALL 2018 1.1. Two-sided z-test. A two-sided test uses a double iterval rejectio regio that icludes both ad. Alterative Hypothesis H A : µ µ 0 x µ 0 σ Critical Values z α/2 ad z α/2 Rejectio Regio (, z α/2 ] or [z α/2, ) P -value 2P [Z > z ] where Z is a stadard ormal distributio. 1.2. The Right-sided z-test. For this test, we use a sigle iterval rejectio regio. Alterative Hypothesis H A : µ > µ 0 x µ 0 σ Critical Value z α Rejectio Regio [z α, ) P -value P [Z > z] where Z is a stadard ormal distributio. 1.3. The Left-sided z-test. For this test, we assume that σ is kow, ad we use a sigle iterval rejectio regio. Alterative Hypothesis H A : µ < µ 0 x µ 0 σ Critical Value z α Rejectio Regio (, z α ] P -value P [Z < z] where Z is a stadard ormal distributio. 2. t-tests for the populatio mea (σ is ukow) For these tests, we assume that σ is ukow. We also assume that is relatively large (> 30) or that the uderlyig populatio is roughly ormal. 2.1. The Two-sided t-test. For this we use a double iterval rejectio regio that icludes both ad. Alterative Hypothesis H A : µ µ 0 x µ 0 s Critical Values t α/2, 1 ad t α/2, 1 Rejectio Regio (, t α/2, 1 ] or [t α/2, 1, )
HYPOTHESIS TESTS FOR ONE POPULATION MEAN WORKSHEET MTH 1210, FALL 2018 3 P -value 2P [T > t ] where T is t-distributio with 1 degrees of freedom. 2.2. The Right-sided t-test. For this test, we use a sigle iterval rejectio regio. Alterative Hypothesis H A : µ > µ 0 x µ 0 s Critical Value t α, 1 Rejectio Regio [t α, 1, ) P -value P [T > t] where T is t-distributio with 1 degrees of freedom. 2.3. The Left-sided t-test. For this test, we use a sigle iterval rejectio regio. Alterative Hypothesis H A : µ < µ 0 x µ 0 s Critical Value t α, 1 Rejectio Regio (, t α, 1 ] P -value P [T < t] where T is t-distributio with 1 degrees of freedom. 3. Exercises For these exercises, use the appropriate test based o the iformatio give (σ, s or ˆp), alterative hypothesis specified ad sigificace level, α. You ca assume that the uderlyig populatio is ormally distributed. (1) x = 300, σ = 37, = 19, H 0 : µ = 320, H A : µ 320, α =.01 (2) x = 300, σ = 37, = 19, H 0 : µ = 320, H A : µ < 320, α =.01 (3) x = 330, σ = 53, = 49, H 0 : µ = 320, H A : µ > 320, α =.1 (4) x = 327, s = 31, = 24, H 0 : µ = 320, H A : µ > 320, α =.1 (5) x = 327, s = 15, = 24, H 0 : µ = 320, H A : µ 320, α =.05 (6) x = 305, s = 23, = 10, H 0 : µ = 320, H A : µ < 320, α =.05 (7) A simple radom sample of 36 St Berard dog weights yields a sample mea of 193 pouds. It is kow that the stadard deviatio of the populatio of all St Berard dog weights is 18 pouds. Test the ull hypothesis that the mea weight of all St Berard dogs is 200 pouds (use a left-sided test ad sigificace level α =.01). State the coclusio i terms of the problem ad calculate the P -value for this test.
4HYPOTHESIS TESTS FOR ONE POPULATION MEAN WORKSHEET MTH 1210, FALL 2018 (8) A simple radom sample of 21 chihuahua dog weights yields a sample mea of 5.6 pouds. It is kow that the stadard deviatio of the populatio of all chihuahua weights is 1.8 pouds. Test the ull hypothesis that the mea weight of all chihuahuas is 4.6 pouds at the α =.1 sigificace level. Use a two-sided test. State the coclusio i terms of the problem ad calculate the P -value for this test. (9) Eleve regios i the Cogolese rai forest are radomly sampled. I each regio raifall was moitored for oe year, ad the followig total yearly raifalls, i cetimeters, were reported: {276, 255, 255, 297, 213, 241, 269, 262, 145, 185, 209} Assume that yearly raifalls withi the Cogolese rai forest are distributed ormally. Test the ull hypothesis that the mea yearly raifall of all locatios i the Cogolese rai forest is 200. Use a two-sided test with α =.01. State the coclusio i terms of the problem. (10) Twelve regios i a Amazo rai forest are radomly sampled. The total yearly raifalls, i cetimeters, for these regios were reported as follows: {188, 232, 210, 198, 202, 193, 219, 202, 252, 156, 184, 222} Assume that yearly raifalls withi the Amazo rai forest are distributed ormally. Test the ull hypothesis that the mea yearly raifall of all locatios i the Amazo rai forest is 200. Use a two-sided test with α =.05. State the coclusio i terms of the problem.
HYPOTHESIS TESTS FOR ONE POPULATION MEAN WORKSHEET MTH 1210, FALL 2018 5 (1) Solutio: 4. Solutios to Exercises 300 320 37 19 = 2.36 Rejectio Regio (, 2.576] or [2.576, ) Coclusio Fail to reject the ull hypothesis at the 1% sigificace level. (2) Solutio: 300 320 37 19 = 2.36 Rejectio Regio (, 2.33] Coclusio Reject the ull hypothesis at the 1% sigificace level. (3) Solutio: 330 320 53 = 1.32 49 Rejectio Regio [1.28, ) Coclusio Reject the ull hypothesis at the 10% sigificace level. (4) Solutio: 327 320 31 = 1.106 24 Rejectio Regio [1.32, ) Coclusio Fail to reject the ull hypothesis at the 10% sigificace level. (5) Solutio: 327 320 15 = 2.29 24 Rejectio Regio (, 2.07] or [2.07, ) Coclusio Reject the ull hypothesis at the 5% sigificace level. (6) Solutio: 305 320 23 10 = 2.06 Rejectio Regio (, 1.83] Coclusio Reject the ull hypothesis at the 5% sigificace level. (7) We have that that = 36, x = 193, ad σ = 18. The critical value of z we eed is z.01 = 2.32 sice we are usig a oe-sided z-test with α =.01. The hypothesis test is the carried out as follows: 193 200 18 = 2.33 36
6HYPOTHESIS TESTS FOR ONE POPULATION MEAN WORKSHEET MTH 1210, FALL 2018 Rejectio Regio (, 2.326] Coclusio Reject the ull hypothesis at the 1% sigificace level. Coclusio i terms of the problem Reject the ull hypothesis that the mea weight of all St Berards is 200 pouds at the 1% level. (8) Solutio: We are give the populatio stadard deviatio, σ, ad our alterative hypothesis is two-sided, so we use the two-sided z-test. 5.6 4.6 1.8 = 2.54 21 Critical Value ±z.05 = ±1.64 Rejectio Regio (, 1.64] or [1.64, ) Coclusio Reject the ull hypothesis at the 10% sigificace level. Coclusio i terms of the problem Reject the ull hypothesis that the true mea weight of all chihuahua dogs is 4.6 pouds at the 10% sigificace level. (9) By iputig the data ito our calculator, we fid that = 11, x = 237, ad s = 44.68. The critical value of t we eed is t.005,10 = 3.17 sice we are usig a two-sided t-test with α =.01 ad 1 = 10. The hypothesis test is the carried out as follows: 237 200 44.68 = 2.74 11 Rejectio Regios (, 3.17] ad [3.17, ) Coclusio Fail to reject the ull hypothesis at the 1% sigificace level. Coclusio i terms of the problem Fail to reject the ull hypothesis that the mea yearly raifall of locatios withi the Cogolese rai forest is 200 cetimeters per year at the 1% level. (10) We are oly give the sample stadard deviatio, s, ad our alterative hypothesis is two-sided, so we use the two-sided t-test. We fid that the mea ad stadard deviatio of this sample are give by x = 204.8 ad s = 24.8, respectively. 204.8 200 24.8 =.67 12 Critical Value ±t.005,11 = ±2.201 Rejectio Regio (, 2.201] or [2.201, ) Coclusio Fail to reject the ull hypothesis at the 5% sigificace level. Coclusio i terms of the problem Fail to reject the ull hypothesis that the true mea yearly raifall at locatios i the Amazo rai forest is 200 iches per year at the 5% sigificace level.