P -values & Rejection Region Tests for a z -test for μ (Large Sample) - PowerPoint Presentation

P-values & Rejection Region Tests for a z-test for μ (Large Sample)n > 30P-values & Rejection Region Tests for a t-test for μ (Small Sample)n < 30Identify H0 and HaIdentify H0 and HaIdentify αIdentify αSTAT Test 1Find pIf p ≤ α, Reject H0If p > α, Fail to reject H0Find zSTAT Test 2Find pIf p ≤ α, Reject H0If p > α, Fail to reject H0Find tFind the critical value(s) and rejection region(s)Use invNorm (2nd VARS 3) functions:Left-tailed test, use αRight-tailed test, use 1 – α Two-tailed test, use Left-tailed test; rejection region to the left of z.Right-tailed test; rejection region to the right of z.Two-tailed test; rejection region is outside of ±z.Find the critical value(s) and rejection region(s)Use invT (2nd VARS 4) functions:Left-tailed test, use αRight-tailed test, use 1 – α Two-tailed test, use Left-tailed test; rejection region to the left of t.Right-tailed test; rejection region to the right of t.Two-tailed test; rejection region is outside of ±t.If standardized test statistic (z) is in the rejection region, reject H0.If standardized test statistic (z) is not in the rejection region, fail to reject H0If standardized test statistic (t) is in the rejection region, reject H0.If standardized test statistic (t) is not in the rejection region, fail to reject H0 P -values & Rejection Region Tests for a z -test for μ (Large Sample) n > 30 P -values & Rejection Region Tests for a t -test for μ (Small Sample) n < 30 Identify H 0 and H a Identify H 0 and H a Identify α Identify α STAT Test 1 Find p If p ≤ α, Reject H 0 If p > α, Fail to reject H 0 Find z STAT Test 2 Find p If p ≤ α, Reject H 0 If p > α, Fail to reject H 0 Find t If standardized test statistic ( z ) is in the rejection region, reject H 0 . If standardized test statistic ( z ) is not in the rejection region, fail to reject H 0 If standardized test statistic ( t ) is in the rejection region, reject H 0 . If standardized test statistic ( t ) is not in the rejection region, fail to reject H 0

The P-value for a hypothesis test is P = 0.0237. What is your decision if the level of significance is (1) α = 0.05 and (2) α = 0.01? 1) Because 0.0237 0.05, you should the null hypothesis. 2) Because 0.0237 0.01, you should the null hypothesis.Find the P-value for a left-tailed hypothesis test with a test statistic of z = -2.23. Decide whether to reject if the level of significance is α = 0.01. For a left-tailed test, the P = (area in left tail), or the area to the left of the given z-score. 2nd Vars 2 (-1E99,-2.23,0,1) = . Since 0.0129 0.01, you should . <Reject>Fail to Reject.0129>Fail to Reject

Find the P-value for a right-tailed hypothesis test with a test statistic of z = 1.987. Decide whether to reject if the level of significance α = 0.05 For a right-tailed test, the P = (area in right tail), or the area to the right of the given z-score. 2nd VARS 2 (1.987, 1E99, 0, 1) = 0.02346 Since P ≤ α, reject .Find the P-value for a two-tailed hypothesis test with a test statistic of z = 2.14. Decide whether to reject if the level of significance α = 0.05 To find P for a two-tailed test, subtract the area between z and -z from 1. 1 – 2nd VARS 2 (-2.14, 2.14, 0, 1) = 0.0324 Since P ≤ α, reject . 

In an advertisement, a pizza shop claims that its mean delivery time is less than 30 minutes. A random selection of 36 delivery times has a sample mean of 28.5 minutes and a standard deviation of 3.5 minutes. Is there enough evidence to support the claim at α = 0.01? Use the critical z-score and rejection region. The claim is “the mean delivery time is less than 30 minutes”. : µ ≥ 30 and : µ < 30 (claim). We are using a z-test because n 30. We will use 30 for µ because that is the claim. We will also use s for , since n 30. We will enter 28.5 for and 3.5 for The critical z-score for the rejection region is found by entering 2nd VARS 3 (0.01) – we use Alpha because it is a left-tail test. The critical z-score is -2.326. Our rejection region is to the LEFT of this value (left tail test). 

In an advertisement, a pizza shop claims that its mean delivery time is less than 30 minutes. A random selection of 36 delivery times has a sample mean of 28.5 minutes and a standard deviation of 3.5 minutes. Is there enough evidence to support the claim at α = 0.01? Use the critical z-score and rejection region. The claim is “the mean delivery time is less than 30 minutes”. : µ ≥ 30 and : µ < 30 (claim). Stat Test 1-Select Stats and enter (30, 3.5, 28.5, 36), select The calculator tells us that the z-score (standardized test statistic) is -2.57. Since this value is to the left of -2.326 (our critical value), it is in the rejection region. We will REJECT the , and SUPPORT the claim. 

In an advertisement, a pizza shop claims that its mean delivery time is less than 30 minutes. A random selection of 36 delivery times has a sample mean of 28.5 minutes and a standard deviation of 3.5 minutes. Is there enough evidence to support the claim at α = 0.01? Use a P-value. The claim is “the mean delivery time is less than 30 minutes”. : µ ≥ 30 and : µ < 30 (claim). We are using a z-test because n 30. We will use 30 for µ because that is the claim. We will also use s for , since n 30. We will enter 28.5 for and 3.5 for Stat Test 1 (30, 3.5, 28.5, 36), select The calculator tells you that p = 0.0051. Since p < α, reject . This means that you support the claim.At the 1% level of significance, you have sufficient evidence to conclude that the mean delivery time is less than 30 minutes. 

Homeowners claim that the mean speed of automobiles traveling on their street is greater than the speed limit of 35 mph. A random sample of 100 automobiles has a mean speed of 36 mph and a standard deviation of 4 mph. Is there enough evidence to support the claim at α = 0.05? Use a P-value. The claim is “the mean speed is greater than 35 mph”. : µ < 35 and : µ > 35 (claim). Stat Test 1 (35, 4, 36, 100); select p = 0.0062. Since p < α, reject .You reject , which means that you support the claim!!At the 5% level of significance, you have sufficient evidence to conclude that the mean automobile speed is greater than 35 mph. 

A study claims that the mean necessary investment to start a franchise business is $143,260. You think that the average franchise investment information is incorrect, so you randomly select 30 franchises and determine the necessary investment for each. The sample mean investment is $135,000 with a standard deviation of $30,000. Is there enough evidence to support your claim at α = 0.05? and (claim). This is a two-tailed test, since the claim is “not equal to”. We are using a z-test (n ≥ 30). We will use 143260 for µ, and 30000 for . Stat Test 1 (143260, 30000, 135000, 30); select The calculator tells you that p = 0.1315. Since p > α, fail to reject .At the 5% level of significance, you do not have sufficient evidence to conclude that the mean necessary franchise investment is different from $143,260. Since you failed to reject , you must fail to support the claim. 

A used car dealership says that the mean price of a 2005 Honda Pilot LX is at least $23,900. You suspect this claim is incorrect and find that a random sample of 14 similar vehicles has a mean price of $23,000 and a standard deviation of $1113. Is there enough evidence to reject the dealer’s claim at ? Assume the population is normally distributed. (claim) This is a left-tailed test with and . Stat Test 2 (n < 30) Select Stats, Enter 23900, 23000, 1113, 14, < You find that your P value is 0.0049. 0.0049 < .05, so you reject . There is enough evidence at the 5% level of significance to reject the claim that the mean price of a 2005 Honda Pilot LX is at least $23,900. 

An industrial company claims that the mean pH level of the water in a nearby river is 6.8. You randomly select 19 water samples and measure the pH of each. The sample mean and standard deviation are 6.7 and 0.24, respectively. Is there enough evidence to reject the company’s claim at ? Assume the population is normally distributed. (claim) This is a two-tail test with and . Stat Test 2 (6.8, 6.7, 0.24, 19); You find that your P value is 0.0860. 0.0860 > .05, so Fail to Reject . There is not enough evidence at the 5% level of significance to reject the claim that the mean pH is 6.8. 

Assignments for 7-2: Classwork: Pages 389-391 #1-14, 29-32 All (Skip #15-28) Homework: Pages 391-393 # 33-45 OddsAssignment for 7-3: Homework: Pages 404-406 # 29-34 All