The test statistic of z=2.58 is obtained when testing the claim that p≠0.587.a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.b.Find the P-value.c. Using a significance level of α=0.05, should we reject H0 or should we fail to reject H0?a.This is a (1)_________b. P-value =_______(Round to three decimal places as needed.)c. Choose the correct conclusion below.A.Reject H0. There is not sufficient evidence to support the claim that p≠0.587.B.Fail to reject H0. There is not sufficient evidence to support the claim that p≠0.587.C.Reject H0. There is sufficient evidence to support the claim that p≠0.587.D.Fail to reject H0. There is sufficient evidence to support the claim that p≠0.587. NEGATIVE z Scores2-3.50 andlower-3.4-3.3-3.2-3.1-3.0-2.9-2.8-2.7-2.6-2.5-24-2.3-22-21-20-1.9-1.8-1.7-1.6-1.5-1.4-1.3-1.2-1.1-1.0-0.9-0.8-0.7-0.6-0.5-0.3-0.2-0.1-0.02Standard Normal (2) Distribution: Cumulative Area from the LEFT.00.01.02.0304.050001.000300050007z Score-1.645-2.575.0010.0013.0019.00260035.0047.00620082.0107.0139017902280287035904460548.0068080809681151.13571587.18412119.242027433085.34463821420746025000.00000300050007.000900130018002500340.05000.00500045.0060006001040136017402220281035104360537065507930961113113351562181420002389270930603409000300050006.000900130018002400330044.0059.0078010201320170021702740344042705260643077809341112.1314.1539.178820612358267630153372NOTE: For values of z below 3.49, use 0.0001 for the area.*Use these common values that result from interpolation:.0003.00040006.00090012.0017.0023.00320043.0067.0075.0099012901660212.0268.033604180516.0630.07640918.1093.12921515.1762203323272643.298133363707409044834880.03.0003.0004.0006.0008001200160023003100410065.0073.0096.0125.0162.0207.0262032904093783.3745416841294562452249604920.01.02Standard Normal (2) Distribution: Cumulative Area from the LEFT.0605.0618.0749.0901.1075.12711492173620052296261129463300366940524443484004.0003.0004.0006.0008.0011.0016.00220030.0040.0054.007100940122.015802020256032204010495080607350885.1056.1251.14691711.19772266.2578.2912.32643632401344044801.05.06.0003.0004.0006.0008.0011.0015.0021.0029.00090062.0069.0091.0119.0154.0197.0250.0314.0392.0485.0594.0721.0869.10381230.1446.16851949.2236254628773228.3594.397443644761.06to.07.00030004.0005.0008.0011.0015.0021.002800:38.0051.0068.0069.011601500192.0244030703840475.0582.07080853.1020.12101423.1660192222062514284331923557.393643254721.07.08.0003.0004.0005.0007.0010.0014.0020.0027.0037.0049.0066.0087.0113.0146.01880239.0301.03750465.0571.0694.0838.1003.11901401.1635.189421772483.2810.31563520.389742864681.0809.0002.0003.0005.0007.0010.0014.0019.0026.0036.0048.0064.0084.0110.0143.01830233.0294.0367.0455.0659.0681.0823.0985.1170.1379.1611.1867.2148.2451.2776.31213483386594247464109-3.50 andlower-3.4-3.3-3.2-3.1-3.0-2.9-2.8-27-2.6-2.5-2.4-2.3-2.2-2.1-2.0-1.9-1.8-1.7-1.6-1.5-1.4-1.3-1.2-1.1-1.0-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-0.0 (1) O two-tailedOright-tailedO left-tailedStandard Normal (2) Distribution: Cumulative Area from the LEFT.04z0.00.1020.30.40.50.60.70.80.91.01.11.21.31.41.51.61.7181.9202.1222.32.42.5262.72.82.93.03.1323.33.50 and up200500053885793.6179.6554691572577580.7881.81598413.8643zScore1.6452.5759032.9192.8332945295549713.97729821.98610803901890389963.9965.997499619967900090059999.00.0150405438583262176991695072917811.79108186843036663809904992079345Area0.95000.99509463356496499719977898269064980698209940.99659966.997598829987.999199939996.999750005478587162556628698573247842.793982128461003690069222935794749556972697839830988898089022.9966996799769902998790019005NOTE: For values of above 3.49, use 0.9999 for the area.*Use these common values that result from interpolation:POSITIVE z Scores03512056175910.6293.701973577673.796782388485.8708.8907.9062.837034841682973297889634.98711001102590439957.99779963996819919006999703.5160.5557.5948.6331.6700706473897704799582648508872989259099926193829495.9591.96719738.9793.9638.9675990499279945.9959.9969.01De04Standard Normal (2) Distribution: Cumulative Area from the LEFT.9977.9984.99889992999499969997.05.6199.5596.5887.6368.6736.70887422.7734802382898531.87498944911592658394.96050600.9678.9744.9798.96429009.9960.9970.9978.9984.9989900290949006.05.065239563660266406.677271237454776480518315877091319279951596089686975098039646968199099931904899619971997999859989999299949996999706.07.6279.5675.6064.6443.6308.71577486.7794.8078.8340.8577.87900900914792929418.9525.9616.9693.9756.9808.9850.9884.9911.99329049.9962.9972.9979.9985.99899992.99959996.9997.0705.5319.57146103.6450.684471907517.7823810683558590.6810.8997.916294290635062556999761.5812.9054.965709131034995199809990190308.5369.5753.6141.6517.68797224.75497852.8133.8389.8621.8630.9015.9177.931994419545.96339706.9767.9817.9667.96900.900.960.999016913691629964.9974.9981.9906.9990919391959997.9998.09Common Critical ValuesConfidence CriticalLevelValue1.6451.962.5750.00.1020.30.40.50.60.70.80.91.01.11.21.31.41.51.61.71.81.9202.12.22.32425262.7282.93.03.13.23.3-3.43.50 and up

a.Suppose p is the true population proportion.

The test statistic of z=2.58 is obtained when testing the claim that p≠0.587. a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed. b.Find the P-value. c. Using a significance level of α=0.05, should we reject H0 or should we fail to reject H0? a.This is a (1)_________ b. P-value =_______ (Round to three decimal places as needed.) c. Choose the correct conclusion below. A.Reject H0. There is not sufficient evidence to support the claim that p≠0.587. B.Fail to reject H0. There is not sufficient evidence to support the claim that p≠0.587. C.Reject H0. There is sufficient evidence to support the claim that p≠0.587. D.Fail to reject H0. There is sufficient evidence to support the claim that p≠0.587.

The test statistic of z=2.58 is obtained when testing the claim that p≠0.587. a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed. b.Find the P-value. c. Using a significance level of α=0.05, should we reject H0 or should we fail to reject H0? a.This is a (1)___ b. P-value =_ (Round to three decimal places as needed.) c. Choose the correct conclusion below. A.Reject H0. There is not sufficient evidence to support the claim that p≠0.587. B.Fail to reject H0. There is not sufficient evidence to support the claim that p≠0.587. C.Reject H0. There is sufficient evidence to support the claim that p≠0.587. D.Fail to reject H0. There is sufficient evidence to support the claim that p≠0.587.

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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