Chapter NineChapter Nine

Inferences Based Inferences Based on Two Sampleson Two Samples

Hypothesis Test 2-Sample Means (Known Hypothesis Test 2-Sample Means (Known ))

Both Normal pdf (known Both Normal pdf (known ))

Null Hypothesis: Null Hypothesis: HH00: : uu11 – – uu2 2 = = 00

Test Statistic: Test Statistic: zz = = x – y – x – y – 00

2211/m + /m + 22

22/n/nAlternative Hypothesis:Alternative Hypothesis: Reject RegionReject Region

HHaa: : uu11 –– uu2 2 > > 00 (Upper Tailed)(Upper Tailed) zz zz

HHaa: : uu11 –– uu2 2 < < 00 ( (Lower Tailed) Lower Tailed) zz - -zz

HHaa: : uu11 –– uu2 2 0 0 (Two-Tailed) either(Two-Tailed) either zz zz/2/2

or or zz - -zz/2/2

P-Value computed the same as 1-Sample Mean.P-Value computed the same as 1-Sample Mean.

Example HT 2-Sample Means (Known Example HT 2-Sample Means (Known ))During a total solar eclipse the temperature drops During a total solar eclipse the temperature drops quickly as the moon passes between the earth and quickly as the moon passes between the earth and the sun. During the June 2001 eclipse in Africa, the sun. During the June 2001 eclipse in Africa, data was collected on the drop in temperature in data was collected on the drop in temperature in degrees F at two types of locations. The average degrees F at two types of locations. The average drop in temperature for 9 samples taken in drop in temperature for 9 samples taken in Mountainous terrain was 15.0. The average drop in Mountainous terrain was 15.0. The average drop in temperature for 12 samples taken in River-level temperature for 12 samples taken in River-level terrain was 17.5. Assume the variance in terrain was 17.5. Assume the variance in temperature drop is known to be 9 for this type of temperature drop is known to be 9 for this type of terrain-temperature drop experiment and that terrain-temperature drop experiment and that experiments of this type follow a Normal pdf. Is experiments of this type follow a Normal pdf. Is there evidence at the there evidence at the = .10 level to conclude that = .10 level to conclude that there is a difference in temperature drop between there is a difference in temperature drop between the two types of terrain in this experiment?P-value?the two types of terrain in this experiment?P-value?

Determining Determining ( ( known) known)

AlternativeAlternative Type II Type IIHypothesisHypothesis Error Error (())HHaa: : uu11 - - uu22 > > 00 zz- - - - 00

HHaa: : uu11 - - uu22 < < 00 1 - 1 - - -zz- - - - 00

HHaa: : uu11 - - uu22 00 zz/2/2 -- - - 00 - - - -zz/2/2- - - - 00

Where Where = = X-YX-Y = = ((2211/m) + (/m) + (22

22/n)/n)

Example HT 2-Sample Means (Known Example HT 2-Sample Means (Known ))A study of report writing by student engineers was A study of report writing by student engineers was conducted at Watson School. A scale that measures the conducted at Watson School. A scale that measures the intelligibility of student engineers’ English is devised. This intelligibility of student engineers’ English is devised. This scale, called an “index of confusion,” is devised, to the scale, called an “index of confusion,” is devised, to the delight of the students, so that low scores indicate high delight of the students, so that low scores indicate high readability. Data are obtained on articles randomly readability. Data are obtained on articles randomly selected from engineering journals and from unpublished selected from engineering journals and from unpublished reports. A sample of 16 engineering journals yielded an reports. A sample of 16 engineering journals yielded an average score of 1.75 while a sample of 25 unpublished average score of 1.75 while a sample of 25 unpublished reports yielded an average score of 2.5. Variance for this reports yielded an average score of 2.5. Variance for this type of scale is known to be 0.48 and the scores are type of scale is known to be 0.48 and the scores are known to follow a Normal pdf. At a significance level known to follow a Normal pdf. At a significance level of .05, does there appear to be a difference between the of .05, does there appear to be a difference between the average scores of the two types of reports? What is the average scores of the two types of reports? What is the Beta error when the true averages differ by as much as Beta error when the true averages differ by as much as 0.5? 0.5?

Hypothesis Test 2-Sample Means (Large n)Hypothesis Test 2-Sample Means (Large n)

Null Hypothesis: Null Hypothesis: HH00: : uu11 – – uu2 2 = = 00

Test Statistic: Test Statistic: zz = = x – y – x – y – 00

ss2211/m + /m + ss22

22/n/nAlternative Hypothesis:Alternative Hypothesis: Reject RegionReject Region

HHaa: : uu11 –– uu2 2 > > 00 (Upper Tailed)(Upper Tailed) zz zz

HHaa: : uu11 –– uu2 2 < < 00 ( (Lower Tailed) Lower Tailed) zz - -zz

HHaa: : uu11 –– uu2 2 0 0 (Two-Tailed) either(Two-Tailed) either zz zz/2/2

or or zz - -zz/2/2

Both m > 40 & n > 40.Both m > 40 & n > 40.

Example HT 2-Sample Means (Large n)Example HT 2-Sample Means (Large n)Aseptic packaging of juices is a method of Aseptic packaging of juices is a method of packaging that entails rapid heating followed by packaging that entails rapid heating followed by quick cooling to room temperature in an air-free quick cooling to room temperature in an air-free container. Such packaging allows the juices to be container. Such packaging allows the juices to be stored un-refrigerated. A new & old machine used stored un-refrigerated. A new & old machine used to fill aseptic packages is being compared. The to fill aseptic packages is being compared. The mean number of containers filled per minute on mean number of containers filled per minute on the new machine was 114.1 for 50 observations the new machine was 114.1 for 50 observations with a standard deviation of 5.0. The mean with a standard deviation of 5.0. The mean number of containers filled per minute on the old number of containers filled per minute on the old machine was 112.7 for 72 observations with a machine was 112.7 for 72 observations with a standard deviation of 3.0. Is there evidence that standard deviation of 3.0. Is there evidence that the new machine is faster than the old machine? the new machine is faster than the old machine? Use a test with Use a test with = .01. (Include the P-value). = .01. (Include the P-value).

Hypothesis Test 2-Sample Means (Small n)Hypothesis Test 2-Sample Means (Small n)

BothBoth Normal pdf (unknown Normal pdf (unknown ))

Null Hypothesis: Null Hypothesis: HH00: : uu11 – – uu2 2 = = 00

Test Statistic: Test Statistic: tt = = x – y – x – y – 00

s s2211/m + s/m + s22

22/n/nAlternative Hypothesis:Alternative Hypothesis: Reject RegionReject Region

HHaa: : uu11 –– uu2 2 > > 00 (Upper Tailed)(Upper Tailed) tt tt, ,

HHaa: : uu11 –– uu2 2 < < 00 ( (Lower Tailed) Lower Tailed) tt - -tt, ,

HHaa: : uu11 –– uu2 2 0 0 (Two-Tailed) either(Two-Tailed) either tt tt/2, /2,

or or tt - -tt/2, /2,

Estimating the Degrees of FreedomEstimating the Degrees of Freedom

ss2211 + + ss22

22 22

= = m n m n (s(s22

11/m)/m)22 + + (s(s2222/n)/n)22

m –1 m –1 n -1 n -1

Example HT 2-Sample Means (Small n)Example HT 2-Sample Means (Small n)The slant shear test is used for evaluating the bond of The slant shear test is used for evaluating the bond of resinous repair materials to concrete. The test utilizes resinous repair materials to concrete. The test utilizes cylinder specimens made of 2 identical halves bonded at cylinder specimens made of 2 identical halves bonded at 303000C. C. Twelve specimens were prepared using wire brushing. The Twelve specimens were prepared using wire brushing. The sample mean shear strength (N/mmsample mean shear strength (N/mm22) and sample standard ) and sample standard deviation were 19.20 & 1.58, respectively. Twelve specimens deviation were 19.20 & 1.58, respectively. Twelve specimens were prepared using hand-chiseled specimens; the were prepared using hand-chiseled specimens; the corresponding values were 23.13 & 4.01.corresponding values were 23.13 & 4.01.Does the true average strength appear to be different for the Does the true average strength appear to be different for the two methods of surface preparation? Use a significance level two methods of surface preparation? Use a significance level of .05 to test the relevant Hypothesis & assume the shear of .05 to test the relevant Hypothesis & assume the shear strength distributions to be Normal. (Include the P-value).strength distributions to be Normal. (Include the P-value).Parameters of interest: R. R.:Parameters of interest: R. R.:Null Hypothesis: Calculation:Null Hypothesis: Calculation:Alternative: Decision: Alternative: Decision: Test Statistic: P-value:Test Statistic: P-value:

Example HT 2-Sample Means (Small n)Example HT 2-Sample Means (Small n)A manufacturer of power-steering components A manufacturer of power-steering components buys hydraulic seals from two sources. Samples buys hydraulic seals from two sources. Samples are selected randomly from among the seals are selected randomly from among the seals obtained from these two suppliers, and each seal obtained from these two suppliers, and each seal is tested to determine the amount of pressure that is tested to determine the amount of pressure that it can withstand. These data result:it can withstand. These data result:Supplier I Supplier I Supplier II Supplier II x = 1342 lb/in x = 1342 lb/in22 y = 1338 lb/in y = 1338 lb/in22

ss22 = 100 = 100 ss22 = 33 = 33 m = 10 m = 10 n = 11n = 11Is there evidence at the Is there evidence at the = .05 level to suggest = .05 level to suggest that the seals from supplier I can withstand higher that the seals from supplier I can withstand higher pressures than those from supplier II? (P-value)pressures than those from supplier II? (P-value)Assume measurements of this type are Normal.Assume measurements of this type are Normal.

HT 2-Sample Means-Paired Data (Small n)HT 2-Sample Means-Paired Data (Small n)

BothBoth Normal X and Y (unknown Normal X and Y (unknown ))

Null Hypothesis: Null Hypothesis: HH00: : uuDD = = 00

Test Statistic: Test Statistic: tt = = dd – – 00

ssD D / / n nAlternative Hypothesis:Alternative Hypothesis: Reject RegionReject RegionHHaa: : uuDD > > 00 (Upper Tailed)(Upper Tailed) tt tt, ,

HHaa: : uuDD < < 00 ( (Lower Tailed) Lower Tailed) tt - -tt, ,

HHaa: : uuDD 0 0 (Two-Tailed) either(Two-Tailed) either tt tt/2, /2,

or or tt - -tt/2, /2,

D = X – Y within each paired observation.D = X – Y within each paired observation.

Example HT 2-Sample Means-Paired Data (Small n)Example HT 2-Sample Means-Paired Data (Small n)One important aspect of computing is the CPU time One important aspect of computing is the CPU time required by an algorithm to solve a problem. A new required by an algorithm to solve a problem. A new algorithm is developed to solve zero-one multiple objective algorithm is developed to solve zero-one multiple objective problems in linear programming. It is thought that a new problems in linear programming. It is thought that a new algorithm will solve problems faster than the algorithm algorithm will solve problems faster than the algorithm currently used. To obtain statistical evidence to support currently used. To obtain statistical evidence to support this research hypothesis, a number of problems will be this research hypothesis, a number of problems will be selected at random. Each problem will be solved twice; selected at random. Each problem will be solved twice; once using the current algorithm and once using the newly once using the current algorithm and once using the newly developed one. These CPU times are not independent; they developed one. These CPU times are not independent; they are based on the same problems solved by two different are based on the same problems solved by two different methods and so are paired by design. The mean difference methods and so are paired by design. The mean difference between the (16) paired data points was 2.7 seconds with a between the (16) paired data points was 2.7 seconds with a standard deviation calculated at 6.0 seconds. Does the data standard deviation calculated at 6.0 seconds. Does the data support this hypothesis at a support this hypothesis at a = .025 level of significance? = .025 level of significance? Assume measurements of this type are known to be Assume measurements of this type are known to be Normal. (Give the P-value) Let X = old & Y = new. Normal. (Give the P-value) Let X = old & Y = new.

Example HT 2-Sample Means-Paired Data (Large n)Example HT 2-Sample Means-Paired Data (Large n)Highway engineers studying the effects of Highway engineers studying the effects of wear on dual-lane highways suspect that wear on dual-lane highways suspect that more cracking occurs in the travel lane of the more cracking occurs in the travel lane of the highway than in the passing lane. To verify highway than in the passing lane. To verify this contention, 64 one-hundred-feet-long test this contention, 64 one-hundred-feet-long test strips are selected, paved, and studied over a strips are selected, paved, and studied over a period of time. It is found that the mean period of time. It is found that the mean difference in the number of major cracks is difference in the number of major cracks is 3.3 with a sample deviation of 8.8. Does this 3.3 with a sample deviation of 8.8. Does this data support the research hypothesis at a data support the research hypothesis at a significance level of .05? (Include P-value). significance level of .05? (Include P-value). Let RV X = Travel lane & RV Y = Passing laneLet RV X = Travel lane & RV Y = Passing lane

The F DistributionThe F Distribution

= = W/W/ = = ++ /2/2 x x((/2)-1/2)-1

Y/Y/ 2 2 x + 1 x + 1 ((++)/2)/2 2 2 2 2

0 < x < 0 < x < W & Y are independent Chi-Square RV’s W & Y are independent Chi-Square RV’s with with & & degrees of freedom. degrees of freedom.

Hypothesis Test on 2 Population VariancesHypothesis Test on 2 Population Variances BothBoth Normal (unknown Normal (unknown uu11 & & uu22))

Null Hypothesis: Null Hypothesis: HH00: : 2211 = = 22

22

Test Statistic: Test Statistic: = = SS221 1 / / SS22

22

Alternative Hypothesis:Alternative Hypothesis: Reject RegionReject Region

HHaa: : 2211 > > 22

22 (Upper Tail)(Upper Tail) FF, m-1, n-1, m-1, n-1

HHaa: : 2211 < < 22

22 (Lower Tail)(Lower Tail) FF1- 1- , m-1, n-1, m-1, n-1

HHaa: : 2211 22

22 (Two-Tail) either(Two-Tail) either FF/2, m-1, n-1/2, m-1, n-1

or or FF1- 1- /2, m-1, n-1/2, m-1, n-1

FF1- 1- , m-1, n-1, m-1, n-1 = 1 / = 1 / FF, n-1, m-1 , n-1, m-1 FF-Tables pg. 730-735-Tables pg. 730-735

Example HT 2-Population VariancesExample HT 2-Population VariancesThe cost of repairing a fiberoptic component may The cost of repairing a fiberoptic component may depend of the stage of production at which it fails. depend of the stage of production at which it fails. The following data are obtained on the cost of The following data are obtained on the cost of repairing parts that fail when installed in the system repairing parts that fail when installed in the system and on the cost of repairing parts that fail after the and on the cost of repairing parts that fail after the system is installed in the field:system is installed in the field: System failureSystem failure Field failureField failure Sample Size = 21 Sample Size = 21 Sample Size = 25 Sample Size = 25 Mean = $65 Mean = $65 Mean = $120 Mean = $120 s s22 = 25 = 25 s s22 = 100 = 100 It is thought that the variance in cost of repairs It is thought that the variance in cost of repairs made in the field is larger than the variance in cost made in the field is larger than the variance in cost of repairs made when the component is placed into of repairs made when the component is placed into the system. Test at the the system. Test at the = .10 level to see if there = .10 level to see if there is statistical evidence to support this contention. is statistical evidence to support this contention.

Example HT 2-Population VariancesExample HT 2-Population VariancesOxide layers on semiconductor wafers are etched Oxide layers on semiconductor wafers are etched in a mixture of gases to achieve the proper in a mixture of gases to achieve the proper thickness. The variability in the thickness of these thickness. The variability in the thickness of these oxide layers is a critical characteristic of the oxide layers is a critical characteristic of the wafer, and low variability is desirable for wafer, and low variability is desirable for subsequent processing steps. Two different subsequent processing steps. Two different mixtures of gases are being studied to determine mixtures of gases are being studied to determine whether one is superior in reducing the variability whether one is superior in reducing the variability of the oxide thickness. Twenty wafers are etched of the oxide thickness. Twenty wafers are etched in each gas. The sample standard deviation of in each gas. The sample standard deviation of oxide thickness are soxide thickness are s11= 1.96 angstroms and s= 1.96 angstroms and s22= =

2.13 angstroms, respectively. Is there any 2.13 angstroms, respectively. Is there any evidence to indicate that either gas is preferable? evidence to indicate that either gas is preferable? Use Use = 0.10 & assume measurements of this type = 0.10 & assume measurements of this type are Normal. are Normal.

Example HT 2-Population VariancesExample HT 2-Population VariancesTwo companies supply raw materials to the Two companies supply raw materials to the manufacturer of paper products. The concentration manufacturer of paper products. The concentration of hardwood in these materials is important for the of hardwood in these materials is important for the tensile strength of the products. The mean tensile strength of the products. The mean concentration of hardwood for both suppliers is the concentration of hardwood for both suppliers is the same; however, the variability in concentration same; however, the variability in concentration may differ between the two companies. The may differ between the two companies. The standard deviation of concentration in a random standard deviation of concentration in a random sample of 26 batches produced by company X is sample of 26 batches produced by company X is 4.7 g/l, while for company Y a random sample of 21 4.7 g/l, while for company Y a random sample of 21 batches yields 6.1 g/l. Is there sufficient evidence batches yields 6.1 g/l. Is there sufficient evidence to conclude that the two population variances to conclude that the two population variances differ? Use differ? Use = .05 & list any assumptions that you = .05 & list any assumptions that you make. make.