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Chapter Eight Tests of Hypothesis Based on a Single Sample

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Hypothesis Testing Elements Null Hypothesis H 0 : > Prior Belief Alternative Hypothesis H a : > Contradictory Belief Test Statistic: > Parameter used to test Rejection Region: > Set of values for rejecting H 0

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Hypothesis Testing Errors Type I Error: > Rejecting H 0 when it is true. Type II Error: > Not rejecting H 0 when it is false.

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Example of Type I Errors Highway engineers have found that many factors affect the performance of reflective highway signs. One is the proper alignment of the car’s headlights. It is thought that more than 50% of the cars on the road have misaimed headlights. If this contention can be supported statistically, then a new tougher inspection program will be put into operation. Let p denote the proportion of cars in operation that have misaimed headlights. Setup a test of hypothesis to test this statement.

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Type II Errors The reject region for the car test is R = 14, 15, …,20 at = Suppose that the true proportion of cars with misaimed headlights is 0.7. What is the probability that our test is unable to detect this situation?

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Hypothesis Testing Protocol Identify parameter of interest State Null & Alternative Hypothesis Give Test Statistic Find Rejection Region for Level Calculate Sample Size for ( ) Decide if H 0 is Rejected or Accepted

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Hypothesis Test about a Population Mean Normal pdf (known ) Null Hypothesis: H 0 : u = u 0 Test Statistic: z = x – u 0 / n Alternative Hypothesis: Reject Region H a : u > u 0 (Upper Tailed) z z H a : u < u 0 (Lower Tailed) z -z H a : u u 0 (Two-Tailed) either z z /2 orz -z /2

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Hypothesis Testing Mean (Known ) Automotive engineers are using more aluminum in the construction of cars in hopes of improving gas mileage. For a particular model the number of miles per gallon obtained currently has a mean of 26.0 mpg with a of 5 mpg. It is hoped that a new design will increase the mean mileage rating. Assume that is not affected by this change. The sample mean for 49 driving tests with this new design yielded mpg. Use a Hypothesis Test with =.05 to make a decision on the validity of this new design to increase the mean mileage rating.

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Hypothesis Testing Mean (Known ) As new engineering manager, you test the melting point of 16 samples of hydrogenated vegetable oil from production, resulting in a sample mean of F. Your company claims a melting point of 95 0 F to all vendors. Using Hypothesis testing, test if your production run meets the 95 0 F specification at level.01. Other evidence indicates a Normal distribution with = 1.20 for the melting point.

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Determining ( known) AlternativeType II Hypothesis Error (u) H a : u > u 0 z + u 0 - u / n H a : u < u -z + u 0 - u / n H a : u u 0 z /2 + u 0 - u - -z /2 + u 0 - u / n / n

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Determining n ( known) n = (z + z ) 2 One Sided u 0 - u n = (z /2 + z ) 2 Two Sided u 0 - u

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Example Type II Error At test level.01, what is the probability of a Type II error when u is actually 94 0 F? What value of n is necessary to ensure that (94) =.10 when =.01?

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Hypothesis Test (Large Sample) Mean Normal pdf (Unknown ) Null Hypothesis: H 0 : u = u 0 Test Statistic: z = x – u 0 s/ n Alternative Hypothesis: Reject Region H a : u > u 0 (Upper Tailed) z z H a : u < u 0 (Lower Tailed) z -z H a : u u 0 (Two-Tailed) either z z /2 orz -z /2 For & n use plausible values for or use Tables A.17

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Hypothesis Test (Large Sample) Mean Ozone is a component of smog that can injure sensitive plants even at low levels. In 1979 a federal ozone standard of 0.12 ppm was set. It is thought that the ozone level in air currents over New England exceeds this level. To verify this contention, air samples are obtained from 64 monitoring stations set up across the region. When the data are analyzed, a sample mean of and a sample SD of 0.03 are obtained. Use a Hypothesis test at a.01 level of significance to test this theory.

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Example HT (Large Sample) Mean The VP of Sales claims that the salesmen are only averaging 15 sales contacts per week. Looking for ways to increase this figure, the VP selects 49 salesmen at random and the number of contacts is recorded for a week. The sample data reveals a mean of 17 contacts with a sample variance of 9. Does the evidence contradict the VP’s claim at the 5% level of significant? Now the VP wants to detect a difference equal to 1 call in the mean number of customer contacts per week. Specifically, he wants to test u = 15 against u = 16. With the same test data, find for this test.

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Hypothesis Test (Small Sample) Mean Normal pdf (Unknown ) Null Hypothesis: H 0 : u = u 0 Test Statistic: t = x – u 0 s/ n Alternative Hypothesis: Reject Region H a : u > u 0 (Upper Tailed) t t ,v H a : u < u 0 (Lower Tailed) t -t ,v H a : u u 0 (Two-Tailed) either t t /2,v or t -t /2,v To find & Sample Size use Table A.17

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Example HT Mean (Small Sample) A new method for measuring phosphorus levels in soil is being tested. A sample of 11 soil specimens with true phosphorus content of 548 mg/kg is analyzed using the new method. The resulting sample mean & sample standard deviation for phosphorus levels are 587 and 10, respectively. Is there evidence that the mean phosphorus level reported by the new method differs significantly from the true value of 548 mg/kg? Use =.05 & assume measurements of this type are Normal.

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Example HT Mean (Small Sample) The true average voltage drop from collector to emitter of insulated gate bipolar transistors is supposed to be at most 2.5 volts. A sample of 10 transistors are used to test if the H 0 : = 2.5 versus H a : > 2.5 volts with =.05. If the standard deviation of the voltage distribution is = 0.10, how likely is it that H 0 will not be rejected when in fact = 2.6?

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Hypothesis Test Population 2 Normal pdf (Unknown ) Null Hypothesis: H 0 : 2 = 0 2 Test Statistic: 2 = (n-1)s 2 0 2 Alternative Hypothesis: Reject Region H a : 2 > 0 2 (Upper Tailed) 2 2 ,v H a : 2 < 0 2 (Lower Tailed) 2 2 1- ,v H a : 2 0 2 (Two-Tailed) either 2 2 /2,v or 2 2 1- /2,v

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Example HT Variance Indoor swimming pools are noted for their poor acoustical properties. The goal is to design a pool in such a way that the average time it takes a low frequency sound to die is at most 1.3 seconds with a standard deviation of at most 0.6 second. Computer simulations of a preliminary design are conducted to see whether these standards are exceeded. The sample mean was 3.97 seconds and the sample standard deviation was 1.89 seconds for 30 simulations. Does it appear that the design specifications are being met at the = 0.01 level for ?

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Example HT Variance A new process for producing small precision parts is being studied. The process consists of mixing fine metal powder with a plastic binder, injecting the mixture into a mold, and then removing the binder with a solvent. Sample data on parts that should have a 1” diameter and whose standard deviation should not exceed inch yielded a sample mean of with a sample standard deviation of for 15 measured parts. Test at the =.05 level to see if this new process is viable for the population .

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P-Values Smallest Level of Significance at which H 0 would be rejected when a specified test procedure is used on a given data set.

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Calculating P-Values One Sided Tests: P = 1 – Φ(z) Upper-tailed P = Φ(z) Lower-tailed Two Sided Test: P = 2 [ 1 – Φ(|z|) ]

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