2One-Sample Tests of Hypothesis GOALSWHAT Define a hypothesis and hypothesis testing.WHY Reasons behind hypothesis testing.HOW Describe the five step hypothesis testing procedure.Distinguish between a one-tailed and a two-tailed test of hypothesis.Conduct a test of hypothesis about a population mean.Define Type I and Type II errors.
3The mean monthly income for systems analysts is $6,325. A statement about the value of a population parameter developed for the purpose of testing.What is aHypothesis?Twenty percent of all customers at Bovine’s Chop House return for another meal within a month.The mean monthly income for systems analysts is $6,325.What is a Hypothesis?
5Hypothesis: A statement about the value of a population parameter developed for the purpose of testing.A particular brand of rice imported to United States contains the arsenic at the level allowable by the EPA.What is a Hypothesis?
6What is Hypothesis Testing? Used to determine whether the hypothesis is a reasonable statement and should not be rejected, or is unreasonable and should be rejectedBased on sample evidence and probability theoryWhat is Hypothesis Testing?
7Why Hypothesis Testing? Because we want to make sure beyond a certain level of doubt and we take into account the sampling errorWhy can’t we just conclude from a sample of rice that has arsenic of 9.5 parts per billionWhy Hypothesis Testing?
9Step One: State the null and alternate hypotheses Null Hypothesis H0A statement about the value of a population parameterAlternative Hypothesis H1:A statement that is accepted if the sample data provide evidence that the null hypothesis is false
103 hypotheses about means Step One: State the null and alternate hypothesesH0: m = 0H1: m = 0Three possibilities regarding meansThe null hypothesis always contains equality.H0: m < 0H1: m > 0H0: m > 0H1: m < 03 hypotheses about means
11Step Two: Select a Level of Significance. Type I ErrorRejecting the null hypothesis when it is actually true (a).Level of SignificanceThe probability of rejecting the null hypothesis when it is actually true; the level of risk in so doing.Type II ErrorAccepting the null hypothesis when it is actually false (b).Step Two: Select a Level of Significance.
12Step Two: Select a Level of Significance. ResearcherNull Accepts RejectsHypothesis Ho HoHo is trueHo is falseCorrectdecisionType I error(a)Type IIError(b)Risk table
13Step Three: Select the test statistic. z Distribution as a test statisticTest statisticA value, determined from sample information, used to determine whether or not to reject the null hypothesis.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).Examples: z, t, F, c2Step Three: Select the test statistic.
14Step 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 DistributionOf the Statistic z, aRight-Tailed Test, .05Level of Significance
15Reject the null hypothesis and accept the alternate hypothesis if Decision RuleReject the null hypothesis and accept the alternate hypothesis ifComputed -z < Critical -zorComputed z > Critical zDecision Rule
16Using the p-Value in Hypothesis Testing 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 testp-ValueDecision RuleIf the p-Value is larger than or equal to the significance level, a, H0 is not rejected.If the p-Value is smaller than the significance level, a, H0 is rejected.Calculated from the probability distribution function or by computerUsing the p-Value in Hypothesis Testing
17Interpreting p-values SOME evidence Ho is not trueVERY STRONG evidence Ho is not trueSTRONG evidence Ho is not true
19One-Tailed Tests of Significance The alternate hypothesis, H1, states a directionH1: The mean yearly commissions earned by full-time realtors is more than $35,000. (µ>$35,000)H1: The mean speed of trucks traveling on I-95 in Georgia is less than 60 miles per hour. (µ<60)H1: Less than 20 percent of the customers pay cash for their gasoline purchase. (p<.20)One-Tailed Tests of Significance
20One-Tailed Test of Significance Sampling DistributionOf the Statistic z, aRight-Tailed Test, .05Level of SignificanceOne-Tailed Test of Significance.
21Two-Tailed Tests of Significance No direction is specified in the alternate hypothesis H1.H1: The mean price for a gallon of gasoline is not equal to $1.54.(µ = $1.54).H1: The mean amount spent by customers at the Wal-mart in Georgetown is not equal to $25.(µ = $25).Two-Tailed Tests of Significance
22Two-Tailed Tests of Significance Regions of Nonrejection and Rejection for a Two-Tailed Test, .05 Level of Significance
23Test for the population mean from a large sample with population standard deviation known Testing for the Population Mean: Large Sample, Population Standard Deviation Known
24The 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 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
25State the decision rule. Reject H0 if z > 1.96 or z < -1.96 Step 4State the decision rule.Reject H0 if z > 1.96or z < -1.96or if p < .05.Step 5Make a decision and interpret the results.Step 3Identify the test statistic. Because we know the population standard deviation, the test statistic is z.Step 1State the null and the alternative hypothesesH0: m = 16H1: m = 16Step 2Select the significance level. The significance level is .05.EXAMPLE 1
26Step 5: Make a decision and interpret the results. The p(z > 1.44) is for a two-tailed test.Computed z of 1.44< Critical z of 1.96,p of > a of .05,Do not reject the null hypothesis.We cannot conclude the mean is different from 16 ounces.Example 1
27As long as the sample size n > 30, z can be approximated using Testing for the Population Mean: Large Sample, Population Standard Deviation UnknownAs long as the sample size n > 30, z can be approximated usingHere s is unknown, so we estimate it with the sample standard deviation s.Testing for the Population Mean: Large Sample, Population Standard Deviation Unknown
28Roder’s Discount Store chain issues its own credit card 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 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
29Make a decision and interpret the results. Step 4H0 is rejected ifz > 1.65or if p < .05.Step 5Make a decision and interpret the results.Step 3Because the sample is large we can use the z distribution as the test statistic.Step 1H0: µ < $400H1: µ > $400Step 2The significance level is .05.Example 2
30Make a decision and interpret the results. Step 5Make a decision and interpret the results.Computed z of 2.42> Critical z of 1.65,p of < a of .05.Reject H0.The p(z > 2.42) is for a one-tailed test.Lisa can conclude that the mean unpaid balance is greater than $400.
31The test statistic is the t distribution. Testing for a Population Mean: Small Sample, Population Standard Deviation UnknownThe test statistic is the t distribution.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
32The current rate for producing 5 amp fuses at Neary Electric Co 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?Example 3
33State the decision rule. There are 10 – 1 = 9 degrees of freedom. The null hypothesis is rejected if t > or, using the p-value, the null hypothesis is rejected if p < .05.Step 4State the decision rule.There are 10 – 1 = 9degrees of freedom.Step 3Find a test statistic. Use the t distribution since s is not known and n < 30.Step 1State the null and alternate hypotheses.H0: µ < 250H1: µ > 250Step 2Select the level of significance. It is .05.
34Make a decision and interpret the results. Step 5Make a decision and interpret the results.The p(t >3.162) is for a one-tailed test.Computed t of 3.162>Critical t ofp of < a of .05Reject HoThe mean number of fuses produced is more than 250 per hour.Example 3
35The fraction or percentage that indicates the part of the population or sample having a particular trait of interest.ProportionThe sample proportion is p and p is the population proportion.Test Statistic for Testing a Single Population Proportion
36In 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
37Find a test statistic. The z distribution is the test statistic. Step 3Find a test statistic. The z distribution is the test statistic.Step 4State the decision rule.The null hypothesis is rejected if z is greater than 1.65 or if p < .05.Step 5Make a decision and interpret the results.Step 1State the null and the alternate hypothesis.H0: p < .15H1: p > .15Step 2Select the level of significance. It is .05.Example 4
38Step 5: Make a decision and interpret the results. p( z > 2.97) =Because the computed z of 2.97 > critical z of 1.65, the p of < a of .05, the null hypothesis is rejected. More than 15 percent responding with a pledge. The new letter is more effective.Example 4
39"Being a statistician means never having to say you are certain.“ A statistician confidently tried to cross a river that was 1 meter deep on average. He drowned."If you torture data enough it will confess"A biologist, a mathematician, and a statistician are on a photo-safari in Africa. They drive out into the savannah in their jeep, stop, and scour the horizon with their binoculars. The biologist: “Look! There’s a herd of zebras! And there, in the middle: a white zebra! It’s fantastic! There are white zebras! We’ll be famous!” The mathematician: “Actually, we know there exists a zebra which is white on one side.” The statistician: “It’s not significant. We only know there’s one white zebra.” The computer scientist: “Oh no! A special case!”