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 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. One-Sample Tests of Hypothesis
10- 4 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? 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.
10- 5 What is Hypothesis Testing? 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
10- 7 Alternative Hypothesis H 1 : A statement that is accepted if the sample data provide evidence that the null hypothesis is false Null Hypothesis H 0 A statement about the value of a population parameter Step One: State the null and alternate hypotheses
10- 8 Three possibilities regarding means H 0 : = 0 H 1 : = 0 H 0 : < 0 H 1 : > 0 H 0 : > 0 H 1 : < 0 Step One: State the null and alternate hypotheses The null hypothesis always contains equality. 3 hypotheses about means
10- 9 Step Two: Select a Level of Significance. The probability of rejecting the null hypothesis when it is actually true; the level of risk in so doing. Rejecting the null hypothesis when it is actually true Accepting the null hypothesis when it is actually false Level of Significance Type I Error Type II Error
10- 10 Step Two: Select a Level of Significance. Researcher Null Accepts Rejects Hypothesis H o H o H o is true H o is false Correct decision Type I error Type II Error Correct decision Risk table
10- 11 Step Three: Select the test statistic. A value, determined from sample information, used to determine whether or not to reject the null hypothesis. Examples: z, t, F, 2 Test statistic z Distribution as a test statistic 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).
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
10- 13 Reject the null hypothesis and accept the alternate hypothesis if Computed -z < Critical -z or Computed z > Critical z Decision Rule
10- 14 Using the p-Value in Hypothesis Testing If the p-Value is larger than or equal to the significance level, , H 0 is not rejected. 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 Calculated from the probability distribution function or by computer Decision Rule If the p-Value is smaller than the significance level, , H 0 is rejected.
10- 15 Interpreting p-values SOME evidence H o is not true STRONG evidence H o is not true VERY STRONG evidence H o is not true
10- 17 One-Tailed Tests of Significance The alternate hypothesis, H 1, states a direction H 1 : The mean yearly commissions earned by full-time realtors is more than $35,000. (µ>$35,000) H 1 : The mean speed of trucks traveling on I- 95 in Georgia is less than 60 miles per hour. (µ<60) H 1 : Less than 20 percent of the customers pay cash for their gasoline purchase. 20)
10- 18 One-Tailed Test of Significance. Sampling Distribution Of the Statistic z, a Right-Tailed Test,.05 Level of Significance
10- 19 Two-Tailed Tests of Significance H 1 : The mean price for a gallon of gasoline is not equal to $1.54. (µ ne $1.54). No direction is specified in the alternate hypothesis H 1. H 1 : The mean amount spent by customers at the Wal-mart in Georgetown is not equal to $25. (µ ne $25). Two-Tailed Tests of Significance
10- 20 Two-Tailed Tests of Significance Regions of Nonrejection and Rejection for a Two- Tailed Test,.05 Level of Significance
10- 21 Testing for the Population Mean: Large Sample, Population Standard Deviation Known Test for the population mean from a large sample with population standard deviation known
10- 22 Example 1 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?
10- 23 EXAMPLE 1 Step 1 State the null and the alternative hypotheses H 0 : = 16 H 1 : 16 Step 3 Identify the test statistic. Because we know the population standard deviation, the test statistic is z. Step 2 Select the significance level. The significance level is.05. Step 4 State the decision rule. Reject H 0 if z > 1.96 or z < -1.96 or if p <.05. Step 5 Make a decision and interpret the results.
10- 24 Example 1 o Computed z of 1.44 < Critical z of 1.96, o p of.1499 > of.05, Do not reject the null hypothesis. The p(z > 1.44) is.1499 for a two-tailed test. Step 5: Make a decision and interpret the results. We cannot conclude the mean is different from 16 ounces.
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
10- 26 Example 2 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?
10- 27 Example 2 Step 1 H 0 : µ < $400 H 1 : µ > $400 Step 2 The significance level is.05. Step 3 Because the sample is large we can use the z distribution as the test statistic. Step 4 H 0 is rejected if z > 1.65 or if p <.05. Step 5 Make a decision and interpret the results.
10- 28 The p(z > 2.42) is.0078 for a one- tailed test. o Computed z of 2.42 > Critical z of 1.65, o p of.0078 < of.05. Reject H 0. Step 5 Make a decision and interpret the results. Lisa can conclude that the mean unpaid balance is greater than $400.
10- 29 Testing for a Population Mean: Small Sample, Population Standard Deviation Unknown 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 test statistic is the t distribution.
10- 30 Example 3 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. At the.05 significance level can Neary conclude that the new machine is faster?
10- 31 Step 4 State the decision rule. There are 10 – 1 = 9 degrees of freedom. Step 1 State the null and alternate hypotheses. H 0 : µ < 250 H 1 : µ > 250 Step 2 Select the level of significance. It is.05. Step 3 Find a test statistic. Use the t distribution since is not known and n < 30. The null hypothesis is rejected if t > 1.833 or, using the p-value, the null hypothesis is rejected if p <.05.
10- 32 Example 3 o Computed t of 3.162 >Critical t of 1.833 o p of.0058 < a of.05 Reject H o The p(t >3.162) is.0058 for a one- tailed test. Step 5 Make a decision and interpret the results. The mean number of amps produced is more than 250 per hour.
10- 33 The sample proportion is p and is the population proportion. The fraction or percentage that indicates the part of the population or sample having a particular trait of interest. Proportion Test Statistic for Testing a Single Population Proportion
10- 34 Example 4 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?
10- 35 Example 4 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. 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.
10- 36 Example 4 Because the computed z of 2.97 > critical z of 1.65, the p of.0015 < of.05, the null hypothesis is rejected. More than 15 percent responding with a pledge. The new letter is more effective. p( z > 2.97) =.0015. Step 5: Make a decision and interpret the results.