Lesson 05 chapter 8 hypothesis testing

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  • Inductive StatisticsDr. Ning [email protected]

    I.007 IBS, Hanze

    Youd better use the full-screen mode to view this PPT file.

    *

  • Table of ContentsReview:Chapter 5 Probability DistributionChapter 6 Sampling DistributionChapter 7 Estimation

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 5: Probability DistributionNormal Distributioncontinuousz=1.00P=0.3413Review:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 5: Probability DistributionNormal Distributioncontinuousz=1.00P=0.6826Review:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 6 Sampling Distribution

    Infinitepopulation Finite populationReview:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 7 EstimationInterval Estimates of the MeanInterval Estimates of the Proportion

    is known: is unknown:

    n

  • Chapter 8 Testing Hypotheses-Summary

    Ch 8 Example P.417H0H1There is no difference between the sample mean and the hypothesized population mean. There is a difference between the sample mean and the hypothesized population mean. H0 : = 10H1 : > 15H1 : < 2H1 : 15For example:MeanReview:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 8 Testing Hypotheses: Practice8-28Step 1: List the known variablesStep 2: Formulate HypothesesStep 3: Calculate the standard errorStep 5: Calculate the z value

    0.05P=0.45z=-1.645

    Step 4: Visualize the confidence levelWith acceptance region accept H0so, new bulb producing is good!Review:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Ch 8 No. Example P.433Example:Step 1: List the known variablesStep 2: Formulate HypothesesStep 3: Calculate the standard errorThe HR director thinks that the average aptitude test is 90. The manager sampled 20 tests and found the mean score is 80 with standard deviation 11.

    If he wants to test the hypothesis at the 0.10 level of significance, what is the procedure?Review:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Ch 8 No. Example P.433Example:Step 4: Visualize the confidence levelStep 5: Calcuate the t valueThe HR director thinks that the average aptitude test is 90. The manager sampled 20 tests and found the mean score is 80 with standard deviation 11.

    If he wants to test the hypothesis at the 0.10 level of significance, what is the procedure?Appendix Table 2

    t=-1.729 +1.729Review:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Ch 8 No. Example P.433Step 4: Visualize the confidence levelAppendix Table 2

    Confidence Intervaldf

    12 0.05 0.10

    1.782Review:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 8 Testing Hypotheses-Summary

    Ch 8 Example P.417H0H1There is no difference between the sample mean and the hypothesized population mean. There is a difference between the sample mean and the hypothesized population mean. H0 : = 10H1 : > 15H1 : < 2H1 : 15For example:MeanProportionReview:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 8 Testing Hypotheses:Proportion

    Ch 8 Example P.427HR director tell the CEO that the promotability of the employees is 80%. The president sampled 150 employees and found that 70% are promotable.

    The CEO wants to test at the 0.05 significance level the hypothesis that 0.8 of the employees are promotable. Example:Step 1: List the known variablesStep 2: Formulate HypothesesStep 3: Calculate the standard errorProportionReview:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 8 Testing Hypotheses:Proportion

    Ch 8 Example P.427HR director tell the CEO that the promotability of the employees is 80%. The president sampled 150 employees and found that 70% are promotable.

    The CEO wants to test at the 0.05 significance level the hypothesis that 0.8 of the employees are promotable. Example:Step 4: Visualize the confidence levelStep 5: Calculate the z score

    ProportionReview:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 8 Testing Hypotheses:Practice

    Ch 8 SC 8-9 P.431. Step 4: Visualize the confidence levelStep 5: Calculate the z scoreStep 1: List the known variablesStep 2: Formulate HypothesesStep 3: Calculate the standard errorSC 8-9Proportion

    Review:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 8 Testing Hypotheses: Measuring Power of a Hypothesis Test

    TrueNot TrueAcceptRejectH0Review:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Test Hypotheses for the MeanTest Hypotheses for the Proportion

    is known is unknown

    n

  • Chapter 8 Testing Hypotheses--Summary

    H0: =XXH1 : > XXH1 : < XXH1 : XX=0.05 z: P=0.45 z= +1.645t: =0.10

    =0.05 z: P=0.475 z= 1.96t: =0.10

    =0.05 z: P=0.45 z= -1.645t: =0.10

    *

  • Chapter 9 Testing Hypotheses: Two-Sample Tests

    Lets compare !Review:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 9 Testing Hypotheses: Two-Sample Tests: Basics

    Independent SamplesDependent SamplesReview:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 9 Testing Hypotheses: Two-Sample Tests: Basics-Independent

    is known: is unknown:

    H0H1

    n

  • Chapter 9 Testing Hypotheses: Two-Sample Tests: Two-Independent Samples

    9.1.1 Difference between means: Large SamplesExample:Ch 9 Example P.456Whether the hourly wages of semiskilled workers are the same between females and males. The survey showed:Step 1: Formulate hypotheses

    Two-tailed TestStep 2: Find the Estimated Standard Error of DifferenceEstimated Standard Error of DifferenceReview:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 9 Testing Hypotheses: Two-Sample Tests : Two-Independent Samples

    z=-1.96 +1.96Step 3: Visualize and find the z valuesStep 2: Find the Standard Error

    Ch 9 Example P.456Review:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 9 Testing Hypotheses: Two-Sample Tests: Two-Independent Samples

    9.1.1 Difference between means: Large SamplesExample:Ch 9 Example P.456Whether the hourly wages of female semiskilled workers are lower than that of males. The survey showed:Step 1: Formulate hypotheses

    One-tailed Test

    P=0.45z=-1.645Step 2: Visualize and find the z values

    Review:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice

    Ch 9 No.9-2 P.460Step 1: Formulate hypotheses

    9-2

    P=0.48 z=Step 2: Find the Standard ErrorStep 3: Visualize and Calculate the z scores

    - 2.05Review:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 9 Testing Hypotheses: Two-Sample Tests: Two-Independent Samples

    9.1.2 Difference between means: Small SamplesExample:Ch 9 Example P.462Which program is more effective in raising sensitivity? The survey showed:Step 1: Formulate hypotheses

    One-tailed Test

    Step 2: Find the Pooled Estimate of 2Review:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 9 Testing Hypotheses: Two-Sample Tests: Two-Independent Samples

    9.1.2 Difference between means: Small SamplesCh 9 Example P.462

    t=1.708

    Step 3: Calculate the standard error

    Step 4: Visualize and find the t scores

    df=(12-1)+(15-1)=25Areas in both tails combined=0.10Review:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice

    Ch 9 No. 9-9 P.466Step 1: Formulate hypotheses

    9-9

    Step 2: Find the Pooled Estimate of 2Step 3: Calculate the standard errorStep 4: Visualize and Find the t scores

    One-tailed Testdf = 16 area=0.10 t=1.746

    Review:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 9 Testing Hypotheses: Two-Sample Tests: Dependent Samples

    9.2 Dependent SamplesReview:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 9 Testing Hypotheses: Two-Sample Tests: Dependent Samples

    9.2 Dependent SamplesCh 9 Example P.468Will the participant lose more than 17 pounds after the weight-reducing program? The survey data is:Step 1: Formulate Hypotheses One-tailed TestExample:Review:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 9 Testing Hypotheses: Two-Sample Tests: Dependent Samples

    9.2 Dependent SamplesCh 9 Example P.468Step 2: Calculate the estimated standard deviation of the population differenceReview:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 9 Testing Hypotheses: Two-Sample Tests: Dependent Samples

    9.2 Dependent SamplesCh 9 Example P.468Step 3: Find the Standard Error of the population differenceStep 4: Calculate the t valueStep 5: Visualize and get the t valuesdf = 10-1=9 area = 0.10t=1.833One-tailed Testreject H0significant differenceReview:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice

    Ch 9 No. 9-15 P.474Step 4: Visalize and Calculate the t valuest=-1.8959-15Step 3: Find the Standard Error of the population differenceStep 1: Formulate Hypotheses Step 2: Calculate the estimated standard deviation of the population differencedf=7 area=0.10reject H0sig differenceReview:Chapter 5 Chapter 6 Chapter 7

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • SummaryReview:Chapter 5 Probability DistributionChapter 6 Sampling DistributionChapter 7 Estimation

    Chapter 8 Testing Hypothesis~Test for Mean* when is known* when is unknown AND n=

  • Connection with BRM(Business Research Methods)

    *

  • Connection with BRM(Business Research Methods)P.354

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  • The Normal Distribution

    SPSS TipsThe data can be downloaded from:

    Blackboard Inductive Statsitics STA2SPSS--Week 5 Correlation and Regression.sav

    *Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.

  • The Normal Distribution

    SPSS TipsOur research data is as below:

    in our research, we are interested in the relationship between the mean response time and the total number correct for 30 puzzles. We obtained scores on 25 adults who are between the ages of 70 and 80 and are not cognitively impaired.

    Please run the SPSS analysis to explore the relationship between the two variables, Latency and Accuracy.

    VariableDescriptionLatencyMean response time for 30 puzzlesAccuracyTotal number correct for 30 puzzles

    *Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.

  • The Normal Distribution

    SPSS TipsStep 1: Click Analyze Correlate Bivariate

    *Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.

  • The Normal Distribution

    SPSS TipsStep 2: Double click on the variables to move to the Variables box

    *Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.

  • The Normal Distribution

    SPSS TipsStep 3: Check it is a two- or one-tailed test and click Options

    *Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.

  • The Normal Distribution

    SPSS TipsStep 4: Click Means and Standard Deviations

    *Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.

  • The Normal Distribution

    SPSS Tips

    *Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.

  • The Normal Distribution

    SPSS TipsStep 5: Click GraphLegacy Dialogs Scatter/Dot...

    *Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.

  • The Normal Distribution

    SPSS Tips

    Step 6: Choose Scatter/Dot Simple scatterplot

    *Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.

  • The Normal Distribution

    SPSS Tips

    Step 7: Choose variables for X,Y axis respectively.

    *Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.

  • The Normal Distribution

    SPSS TipsNow you know something about the correlation, but how can you get the regression line as below?

    *Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.

  • The Normal Distribution

    SPSS TipsThe correlation between latency and accuracy is -.545, indicating the greater the latency the less the accuracy. The p value of .005 indicates we reject at the .05 level the null hypothesis that latency and accuracy are linearly unrelated in the population.

    An examination of the bivariate scatterplot supports the conclusion that there is a fairly strong negative linear relationship between the two variables. Interpretation:

    *Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.

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    *Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.*Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.*Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.*Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.*Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.*Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.*Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.*Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.*Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.*Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.*Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.*Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y, which equals 10 X, is a binomial random variable with n = 10 and q = 1 p = 0.30. We are interested in finding P(X 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewriteP(X 4) as a probability statement in terms of Y:P(X 4) = P(X 4) = P(10 X 10 4) = P(Y 6)Now it's just a matter of looking up the probability in the right place on our cumulative binomial table.To findP(Y 6), we:Findn= 10in the first column on the left.Find the column containingp= 0.30.Find the6in the second column on the left, since we want to findF(6) =P(Y 6).Now, all we need to do is read the probability value where thep= 0.30 column and the (n= 10, y= 6) row intersect.What do you get?Do you need a hint?

    The cumulative binomial probability table tells us thatP(Y 6)=P(X 4) = 0.9894.That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.