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10- 1 Chapter Ten McGraw-Hill/Irwin © 2005 The McGraw-Hill Companies, Inc., All Rights Reserved. 10- 2 Chapter Ten One-Sample Tests of Hypothesis GOALS When you have completed this chapter, you will be able to: ONE Define a hypothesis and hypothesis testing. TWO Describe the five step hypothesis testing procedure. THREE Distinguish between a one-tailed and a two-tailed test of hypothesis. FOUR Conduct a test of hypothesis about a population mean. 10- 3 Chapter Ten continued One-Sample Tests of Hypothesis GOALS When you have completed this chapter, you will be able to: FIVE Conduct a test of hypothesis about a population proportion. SIX Define Type I and Type II errors. SEVEN Compute the probability of a Type II error. 10- 4 A statement about the value of a population parameter developed for the purpose of testing. The mean monthly income for systems analysts is $6,325. What is a Hypothesis? Twenty percent of all customers at Bovine’s Chop House return for another meal within a month. What is a Hypothesis? 10- 5 Hypothesis testing Based on sample evidence and probability theory Used to determine whether the hypothesis is a reasonable statement and should not be rejected, or is unreasonable and should be rejected What is Hypothesis Testing? 10- 6 Step 1: State null and alternate hypotheses Step 2: Select a level of significance Step 3: Identify the test statistic Step 4: Formulate a decision rule Step 5: Take a sample, arrive at a decision Do not reject null Reject null and accept alternate Hypothesis Testing 10- 7 Step One: State the null and alternate hypotheses Null Hypothesis H0 A statement about the value of a population parameter Alternative Hypothesis H1: A statement that is accepted if the sample data provide evidence that the null hypothesis is false 10- 8 Step One: State the null and alternate hypotheses Three possibilities regarding means H 0: m = 0 H 1: m = 0 H 0: m < 0 H 1: m > 0 H0: m > 0 H1: m < 0 The null hypothesis always contains equality. 3 hypotheses about means 10- 9 Type I Error Level of Significance Rejecting the null hypothesis when it is actually true (a). The probability of rejecting the null hypothesis when it is actually true; the level of risk in so doing. Type II Error Accepting the null hypothesis when it is actually false (b). Step Two: Select a Level of Significance. 10- 10 Step Two: Select a Level of Significance. Null Hypothesis Ho is true Ho is false Researcher Accepts Rejects Ho Ho Correct Type I error decision (a) Type II Correct Error decision (b) Risk table 10- 11 Test statistic A value, determined from sample information, used to determine whether or not to reject the null hypothesis. Examples: z, t, F, c2 z Distribution as a test statistic X m z / n The z value is based on the sampling distribution of X, which is normally distributed when the sample is reasonably large (recall Central Limit Theorem). Step Three: Select the test statistic. 10- 12 Step Four: Formulate the decision rule. Critical value: The dividing point between the region where the null hypothesis is rejected and the region where it is not rejected. Sampling Distribution Of the Statistic z, a Right-Tailed Test, .05 Level of Significance Region of Do not rejection reject [Probability =.95] 0 [Probability=.05] 1.65 Critical value 10- 13 Decision Rule Reject the null hypothesis and accept the alternate hypothesis if Computed -z < Critical -z or Computed z > Critical z Decision Rule 10- 14 p-Value The probability, assuming that the null hypothesis is true, of finding a value of the test statistic at least as extreme as the computed value for the test Decision Rule If the p-Value is larger If the p-Value is than or equal to the smaller than the significance level, a, H0 significance level, a, is not rejected. H0 is rejected. Calculated from the probability distribution function or by computer Using the p-Value in Hypothesis Testing 10- 15 Interpreting p-values >.05 p >.001 p .10 SOME evidence Ho is not true >.01 p STRONG evidence Ho is not true .05 .01 VERY STRONG evidence Ho is not true 10- 16 Step Five: Make a decision. Movie 10- 17 One-Tailed Tests of Significance The alternate hypothesis, H1, states a direction H1: The mean speed of trucks traveling on I95 in Georgia is less than 60 miles per hour. (µ<60) H1: The mean yearly commissions earned by full-time realtors is more than $35,000. (µ>$35,000) H1: Less than 20 percent of the customers pay cash for their gasoline purchase. (p<.20) One-Tailed Tests of Significance 10- 18 Sampling Distribution Of the Statistic z, a Right-Tailed Test, .05 Level of Significance Region of Do not rejection reject [Probability =.95] 0 [Probability=.05] 1.65 Critical value One-Tailed Test of Significance . 10- 19 Two-Tailed Tests of Significance No direction is specified in the alternate hypothesis H1. H1: The mean amount spent by customers at the Wal-mart in Georgetown is not equal to $25. (µ ne $25). H1: The mean price for a gallon of gasoline is not equal to $1.54. (µ ne $1.54). Two-Tailed Tests of Significance 10- 20 Two-Tailed Tests of Significance Regions of Nonrejection and Rejection for a TwoTailed Test, .05 Level of Significance Region of Region of Do not rejection rejection reject [Probability=.025] [Probability =.95] -1.96 Critical value 0 [Probability=.025] 1.96 Critical value 10- 21 Test for the population mean from a large sample with population standard deviation known X m z / n Testing for the Population Mean: Large Sample, Population Standard Deviation Known 10- 22 The processors of Fries’ Catsup indicate on the label that the bottle contains 16 ounces of catsup. The standard deviation of the process is 0.5 ounces. A sample of 36 bottles from last hour’s production revealed a mean weight of 16.12 ounces per bottle. At the .05 significance level is the process out of control? That is, can we conclude that the mean amount per bottle is different from 16 ounces? Example 1 10- 23 Step 5 Make a decision and interpret the results. Step 4 State the decision rule. Reject H0 if z > 1.96 or z < -1.96 or if p < .05. Step 3 Identify the test statistic. Because we know the population standard deviation, the test statistic is z. Step 1 State the null and the alternative hypotheses H0: m = 16 H1: m 16 Step 2 Select the significance level. The significance level is .05. EXAMPLE 1 10- 24 Step 5: Make a decision and interpret the results. z X m n oComputed 16 .12 16 .00 0.5 z of 1.44 < Critical z of 1.96, op of .1499 > a of .05, Do not reject the null hypothesis. 36 1.44 The p(z > 1.44) is .1499 for a two-tailed test. We cannot conclude the mean is different from 16 ounces. Example 1 10- 25 Testing for the Population Mean: Large Sample, Population Standard Deviation Unknown Here is unknown, so we estimate it with the sample standard deviation s. As long as the sample size n > 30, z can be approximated using X m z s/ n Testing for the Population Mean: Large Sample, Population Standard Deviation Unknown 10- 26 Roder’s Discount Store chain issues its own credit card. Lisa, the credit manager, wants to find out if the mean monthly unpaid balance is more than $400. The level of significance is set at .05. A random check of 172 unpaid balances revealed the sample mean to be $407 and the sample standard deviation to be $38. Should Lisa conclude that the population mean is greater than $400, or is it reasonable to assume that the difference of $7 ($407$400) is due to chance? Example 2 10- 27 Step 5 Make a decision and interpret the results. Step 4 H0 is rejected if z > 1.65 or if p < .05. Step 3 Because the sample is large we can use the z distribution as the test statistic. Step 1 H0: µ < $400 H1: µ > $400 Step 2 The significance level is .05. Example 2 Step 5 Make a decision and interpret the results. 10- 28 z oComputed z of 2.42 > Critical z of 1.65, op of .0078 < a of .05. Reject H0. Lisa can conclude that the mean unpaid balance is greater than $400. X m s n $407 $400 $38 2.42 172 The p(z > 2.42) is .0078 for a onetailed test. 10- 29 Testing for a Population Mean: Small Sample, Population Standard Deviation Unknown The test statistic is the t distribution. t X m s/ n The critical value of t is determined by its degrees of freedom equal to n-1. Testing for a Population Mean: Small Sample, Population Standard Deviation Unknown 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. The production hours are normally distributed. A sample of 10 randomly selected hours from last month revealed that the mean hourly production on the new machine was 256 units, with a sample standard deviation of 6 per hour. 10- 30 At the .05 significance level can Neary conclude that the new machine is faster? Example 3 10- 31 The null hypothesis is rejected if t > 1.833 or, using the p-value, the null hypothesis is rejected if p < .05. Step 4 State the decision rule. There are 10 – 1 = 9 degrees of freedom. Step 1 State the null and alternate hypotheses. H0: µ < 250 H1: µ > 250 Step 3 Find a test statistic. Use the t distribution since is not known and n < 30. Step 2 Select the level of significance. It is .05. 10- 32 Step 5 Make a decision and interpret the results. t X m s n 256 250 6 10 3.162 The p(t >3.162) is .0058 for a onetailed test. oComputed t of 3.162 >Critical t of 1.833 op of .0058 < a of .05 Reject Ho The mean number of amps produced is more than 250 per hour. Example 3 10- 33 Proportion The sample proportion is p and p is the population proportion. The fraction or percentage that indicates the part of the population or sample having a particular trait of interest. Number of successesin the sample p Number sampled Test Statistic for Testing a Single Population Proportion z p p p (1 p ) n 10- 34 In the past, 15% of the mail order solicitations for a certain charity resulted in a financial contribution. A new solicitation letter that has been drafted is sent to a sample of 200 people and 45 responded with a contribution. At the .05 significance level can it be concluded that the new letter is more effective? Example 4 10- 35 Step 3 Find a test statistic. The z distribution is the test statistic. Step 4 State the decision rule. The null hypothesis is rejected if z is greater than 1.65 or if p < .05. Step 5 Make a decision and interpret the results. Step 1 State the null and the alternate hypothesis. H0: p < .15 H1: p > .15 Step 2 Select the level of significance. It is .05. Example 4 10- 36 Step 5: Make a decision and interpret the results. 45 .15 p p z 200 2.97 p (1 p ) .15 (1 .15 ) n 200 p( z > 2.97) = .0015. Because the computed z of 2.97 > critical z of 1.65, the p of .0015 < a of .05, the null hypothesis is rejected. More than 15 percent responding with a pledge. The new letter is more effective. Example 4