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One Sample Tests of Hypothesis Chapter 10 McGraw-Hill/Irwin Copyright © 2012 by The McGraw-Hill Companies, Inc. All rights reserved. Learning Objectives Chapter 10 1 Define a hypothesis. 3 Describe Type I and Type II errors. 5 Distinguish between a one-tailed and two-tailed hypothesis 6 Apply the five-step procedure to the test of hypothesis about a population mean. 7 Compute and interpret a p-value. 8 Conduct a test of hypothesis about a population proportion. 9-2 Hypothesis and Hypothesis Testing HYPOTHESIS A statement about the value of a population parameter developed for the purpose of testing. HYPOTHESIS TESTING A procedure based on sample evidence and probability theory to determine whether the hypothesis is a reasonable statement. 10-3 Null and Alternate Hypothesis NULL HYPOTHESIS A statement about the value of a population parameter developed for the purpose of testing numerical evidence. ALTERNATE HYPOTHESIS A statement that is accepted if the sample data provide sufficient evidence that the null hypothesis is false. 10-4 Hypotheses: H0 and H1 H0: null hypothesis and H1: alternate hypothesis Examples: H0: defendant is innocent; H1: defendant is guilty H0: µ =10; H1: µ >10 H0 is presumed to be true, but in business analysis we usually seek to reject it, that is, to prove it is not true. H1 has the burden of proof We collect evidence from the sample. If our decision is 'do not reject H0', this does not necessarily mean that the null hypothesis is true, it only suggests that there is not sufficient evidence to reject H0; rejecting the null hypothesis then, suggests that the alternative hypothesis may be true. Equality is used in H0; “≠” “<” and “>” is used in H1 10-5 Decisions and Errors in Hypothesis Testing (Acquaint) (Defendant innocent) Correct decision I error (Defendant Type Type II error guilty) P(Type P(Type IIII error)=β error)=β (Convict) Type I error P(Type I error)=α P(Type I Also called error)=α significance level Correct decision 10-6 Test Statistic versus Critical Value TEST STATISTIC A value, determined from sample information, used to determine whether to reject the null hypothesis. Example: z, t, F, 2 CRITICAL VALUE The dividing point between the region where the null hypothesis is rejected and the region where it is not rejected. 10-7 One-tail vs. Two-tail Test— H :<A depending on H1 1 H1: ≠ A H1: > A 10-8 Hypothesis Setups for Testing a Population Mean () Required condition: the data is approximately normally distributed. Two-tail test Left-tail test = Right-tail test = Z > Zα/2 or Z < Zα/2 t > tα/2, n-1 or t < tα/2, n-1 With σ unknown: With σ known: 10-9 Testing for a Population Mean with a Known Population Standard Deviation- Example Jamestown Steel Company manufactures and assembles desks and other office equipment . The weekly production of the Model A325 desk at the Fredonia Plant follows the normal probability distribution with a mean of 200 and a standard deviation of 16. Recently, new production methods have been introduced and new employees hired. The VP of manufacturing would like to investigate whether there has been a change in the weekly production of the Model A325 desk. In other words, is the mean number of desks produced at the Fedonia Plant different from 200 at the .01 significance level? 10-10 Testing for a Population Mean with a Known Population Standard Deviation- Example To investigate this issue, the weekly production of 50 weeks from last year was obtained (the plant was shut down 2 weeks for vacation). The sample mean is 203.5. What we know from the description: 16, X 203.5, n 50 , 0.01 10-11 Testing for a Population Mean with a Known Population Standard Deviation- Example Step 1: State the null hypothesis and the alternate hypothesis. H0: = 200 H1: ≠ 200 (note: keyword in the problem “has changed”) Step 2: Select the level of significance. α = 0.01 as stated in the problem Step 3: Select the test statistic. Use Z-distribution since σ is known 10-12 Testing for a Population Mean with a Known Population Standard Deviation- Example Step 3: Select the test statistic. Use Z-distribution since σ is known Z 203.5 200 1.55 16 / 50 10-13 Testing for a Population Mean with a Known Population Standard Deviation- Example Step 4: Formulate the decision rule. Rejection Region: Reject H0 if |Z| > Z/2 Z Z / 2 or Z Z / 2 Z / 2 Z.01/ 2 2.58 Rejection H0 if Z 2.58 or Z -2.58 =NORMSINV(1-0.005) Step 5: Make a decision and interpret the result. Because 1.55 does not fall in the rejection region, H0 is not rejected. We conclude that the population mean is not different from 200. So we would report to the vice president of manufacturing that the sample evidence does not show that the production rate at the plant has changed from 200 per week. 10-14 Testing for a Population Mean with a Known Population Standard Deviation- Another Example Suppose in the previous problem the vice president wants to know whether there has been an increase in the number of units assembled. To put it another way, can we conclude, because of the improved production methods, that the mean number of desks assembled is more than 200? Recall: σ=16, n=50, α=.01 10-15 Testing for a Population Mean with a Known Population Standard Deviation- Example Step 1: State the null hypothesis and the alternate hypothesis. H0: = 200 H1: > 200 (note: keyword in the problem “an increase”) Step 2: Select the level of significance. α = 0.01 as stated in the problem Step 3: Select the test statistic. Use Z-distribution since σ is known Z 203 .5 200 1.55 16 / 50 10-16 One-Tailed Test versus Two-Tailed Test = 10-17 Testing for a Population Mean with a Known Population Standard Deviation- Example Step 4: Formulate the decision rule. Reject H0 if Z > Z Z Z Z 0.01 2.33 Rejection H 0 if Z 2.33 1.55 Step 5: Make a decision and interpret the result. Because 1.55 does not fall in the rejection region, H0 is not rejected. We conclude that the average number of desks assembled in the last 50 weeks is not more than 200 10-18 Testing for the Population Mean: Population Standard Deviation Unknown When the population standard deviation (σ) is unknown, the sample standard deviation (s) is used in its place The t-distribution is used as test statistic, which is computed using the formula: 10-19 Characteristics of the t-distribution 1. It is a continuous distribution, bell-shaped and symmetrical. 2. There is a family of t distributions. The shape of the t distribution is controlled by degrees of freedom, denoted by ν, where ν = n – 1. 3. All t distributions have a mean of 0, but their standard deviations differ according to the sample size, n. 4. The t distribution is more spread out and flatter at the center than the standard normal distribution. 5. As the sample size increases, the t distribution approaches the standard normal distribution; when the degrees of freedom is ∞, t distribution becomes standard normal distribution. 9-20 Testing for the Population Mean: Population Standard Deviation Unknown - Example The McFarland Insurance Company Claims Department reports the mean cost to process a claim is $60. An industry comparison showed this amount to be larger than most other insurance companies, so the company instituted cost-cutting measures. To evaluate the effect of the costcutting measures, the Supervisor of the Claims Department selected a random sample of 26 claims processed last month. The sample information is reported below. At the .01 significance level is it reasonable a claim is now less than $60? Data McFarland Insurance Claim 10-21 Testing for a Population Mean with a Known Population Standard Deviation- Example Step 1: State the null hypothesis and the alternate hypothesis. H0: = $60 H1: < $60 (note: keyword in the problem “now less than”) Step 2: Select the level of significance. α = 0.01 as stated in the problem Step 3: Select the test statistic. Use t-distribution since σ is unknown t 56.42 60 1.818 10.04 / 26 10-22 Testing for a Population Mean with a Known Population Standard Deviation- Example Step 4: Formulate the decision rule. Reject H0 if t < -t,n-1 t t ,n 1 t0.01, 261 2.485 Rejection H 0 if t -2.485 =TINV(2α, d.f.)=TINV(0.02, 25) Step 5: Make a decision and interpret the result. Because -1.818 does not fall in the rejection region, H0 is not rejected at the .01 significance level. We have not demonstrated that the cost-cutting measures reduced the mean cost per claim to less than $60. The difference of $3.58 ($56.42 - $60) between the sample mean and the population mean could be due to sampling error. 10-23 Testing for a Population Mean with an Unknown Population Standard Deviation- Example The current rate for producing 5 amp fuses at Neary Electric Co. is 250 per hour. A new machine has been purchased and installed that, according to the supplier, will increase the production rate. A sample of 10 randomly selected hours from last month revealed the mean hourly production on the new machine was 256 units, with a sample standard deviation of 6 per hour. At the .05 significance level can Neary conclude that the new machine is faster? 10-24 Testing for a Population Mean with an Unknown Population Standard Deviation- Example Step 1: State the null and the alternate hypothesis. H0: µ = 250 H1: µ > 250 Step 2: Select the level of significance. It is .05. Step 3: Find a test statistic. X 256 250 t 3.162 s/ n 6 / 10 10-25 Testing for a Population Mean with an Unknown Population Standard Deviation- Example Step 4: State the decision rule. There are 10 – 1 = 9 degrees of freedom. The null hypothesis is rejected if t > 1.833. t t ,n 1 t0.05,101 1.833 Step 5: Make a decision and interpret the results. The null hypothesis is rejected. There is enough evidence in support of the alternative hypothesis. The mean number produced is more than 250 per hour. 10-26 p-Value in Hypothesis Testing p-VALUE is the probability of observing a sample value as extreme as, or more extreme than, the value observed, given that the null hypothesis is true. In testing a hypothesis, we can also compare the pvalue to the significance level (). Decision rule using the p-value: Reject H0 if p-value < significance level 10-27 p-Value in Hypothesis Testing - Example Recall the last problem where the hypothesis and decision rules were set up as: H0: = 200 H1: > 200 Reject H0 if Z > 2.33, where Z = 1.55 P-value: What is the probability of observing a sample mean as high as 203.5 when the population mean is 200 and the population standard deviation is 16? P-value=P(Z>1.55)=P(Z<-1.55)=.0606 =NORMDIST(-1.55) Reject H0 if p-value < , where =0.01 Decision: Do not reject H0 10-28 What does it mean when p-value < ? (a) .10, we have some evidence that H0 is not true. (b) .05, we have strong evidence that H0 is not true. (c) .01, we have very strong evidence that H0 is not true. (d) .001, we have extremely strong evidence that H0 is not true. 10-29 P-value—two-tail test example Two-tail test H0: = 200 H1: ≠ 200 Test statistic: Z 1.55 P-value: 2P(Z<neg test stat)=2P(Z<-1.55)=2(.0606)=.1212 P-value>0.01 Decision: do not reject the null hypothesis Summary Two-tail test Left-tail test = p-value=2P(Z<neg test stat) p-value=P(Z<test stat) p-value=2P(t<neg test stat) p-value=P(t<test stat) Right-tail test = p-value=P(Z>test stat) p-value=P(t>test stat) P-value for t-test: =tdist(| t |, d.f., tail), where tail=1 for one-tail test and tail=2 for two-tail test. Assumptions in Testing a Population Proportion using the z-Distribution A random sample is chosen from the population. the sample data collected are the result of counts. the outcome of an experiment is classified into one of two mutually exclusive categories—a “success” or a “failure.” “Success” refers to the category we want to study; “failure” refers to any other categories. Both n and n(1- ) are at least 5. When the above conditions are met, the normal distribution can be used as an approximation to the binomial distribution. 10-32 Hypothesis Setups for Testing a Proportion () Z > Zα/2 or Z < Zα/2 10-33 Test Statistic for Testing a Single Population Proportion Hypothesized population proportion Sample proportion z p (1 ) n Sample size 10-34 Test Statistic for Testing a Single Population Proportion - Example The GO transportation system of buses and commuter trains operates on the honor system. Train travelers are expected to buy their tickets before boarding the train. Only a small number of people will be checked on the train to see whether they bought a ticket. Suppose that a random sample of 400 train travelers was sampled and 68 of them had failed to buy a ticket. Test to determine whether the proportion of travelers who fail to buy a ticket exceeds 15%. Use a 5% significance level. 10-35 Test Statistic for Testing a Single Population Proportion - Example Step 1: State the null hypothesis and the alternate hypothesis. H0: = .15 H1: > .15 (note: keyword in the problem “exceeds”) Step 2: Select the level of significance. α = 0.05 as stated in the problem Step 3: Select the test statistic. Use Z-distribution since the assumptions are met. 68 p .17 .15 p .17 Z 1.12 400 (1 ) .15(1 .15) n 400 10-36 Testing for a Population Proportion Example Step 4: Formulate the decision rule. Reject H0 if Z < Z Zα = Z.05 = 1.645 Rejection region: reject H0 if Z > 1.645 Z = 1.12 Step 5: Make a decision and interpret the result. The test statistic (Z=1.12) is not in the rejection region, so the null hypothesis is not rejected at the .05 level. This indicates that there is not enough evidence to infer the prportion of travelers who fail to buy a ticket exceeds 15%. 10-37 Testing for a Population Proportion - P-value Hypotheses H0: = .15 H1: > .15 Test statistic z=1.12 P-value P(Z > test stat)=P(Z > 1.12)=P(Z<-1.12)=0.1314 P-value>0.05 Decision Do not reject the null hypothesis

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