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1 ES9 Chapter 9 ~ Inferences Involving One Population Students t, df = 5 Students t, df = 15 Students t, df = 25.

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Presentation on theme: "1 ES9 Chapter 9 ~ Inferences Involving One Population Students t, df = 5 Students t, df = 15 Students t, df = 25."— Presentation transcript:

1 1 ES9 Chapter 9 ~ Inferences Involving One Population Students t, df = 5 Students t, df = 15 Students t, df = 25

2 2 ES9 Chapter Goals Learned about confidence intervals and hypothesis testing Assumed was known Consider both types of inference about when is unknown Consider both types of inference about p, the binomial probability of success Consider the inference of hypothesis testing of, the population standard deviation

3 3 ES9 Inferences about are based on the sample mean If the sample size is large or the sample population is normal: has a standard normal distribution )//()(*nxz 9.1 ~ Inference About mean ( unknown) If is unknown, use s as a point estimate for Estimated standard error of the mean: Test statistic:

4 4 ES9 1.When s is used as an estimate for, the test statistic has two sources of variation: sx and 4.The population standard deviation,, is almost never known in real-world problems The standard error will almost always be estimated using Almost all real-world inference about the population mean will be completed using the Students t-statistic Students t-Statistic 3.Assumption: samples are taken from normal populations 2.The resulting test statistic: Known as the Students t-statistic

5 5 ES9 Properties of the t-Distribution (df>2) 3.t is distributed so as to form a family of distributions, a separate distribution for each different number of degrees of freedom 1.t is distributed with a mean of 0 2.t is distributed symmetrically about its mean 4.The t-distribution approaches the normal distribution as the number of degrees of freedom increases 5.t is distributed with a variance greater than 1, but as the degrees of freedom increase, the variance approaches 1 6.t is distributed so as to be less peaked at the mean and thicker at the tails than the normal distribution

6 6 ES9 Students t, df = 5 Students t, df = 15 Normal distribution Degrees of Freedom, df: A parameter that identifies each different distribution of Students t-distribution. For the methods presented in this chapter, the value of df will be the sample size minus 1, df = n 1. Students t-Distributions

7 7 ES9 1.The number of degrees of freedom associated with s 2 is the divisor (n 1) used to define the sample variance s 2 Thus: df = n 1 Notes 2.The number of degrees of freedom is the number of unrelated deviations available for use in estimating 2 3.Table for Students t-distribution (Table 6 in Appendix B) is a table of critical values. Left column = df. When df > 100, critical values of the t-distribution are the same as the corresponding critical values of the standard normal distribution. 4.Notation: t (df, ) Read as: t of df,

8 8 ES9 ) df, ( t t-Distribution Showing t (df, )

9 9 ES9 Example Example:Find the value of t (12, 0.025) Portion of Table t (12, 0.025) t (12, 0.025)

10 10 ES9 1.If the df is not listed in the left-hand column of Table 6, use the next smaller value of df that is listed Notes cumulative probability 2.Most computer software packages will calculate either the area related to a specified t-value or the t-value that bounds a specified area 3.The cumulative distribution function (CDF) is often used to find area from to t 4.If the area from to t is known and the value of t is wanted, then the inverse cumulative distribution function (INVCDF) is used

11 11 ES9 Confidence Interval Procedure: 1.Procedure for constructing confidence intervals similar to that used when is known The assumption for inferences about mean when is unknown: The sampled population is normally distributed The Assumption... 2.Use t in place of z, use s in place of 3.The formula for the confidence interval for is: 1df wheret (df, to n n s x n s xt (df,

12 12 ES9 Example:A study is conducted to learn how long it takes the typical tax payer to complete their federal income tax return. A random sample of 17 income tax filers showed a mean time (in hours) of 7.8 and a standard deviation of 2.3. Find a 95% confidence interval for the true mean time required to complete a federal income tax return. Assume the time to complete the return is normally distributed. Solution: 1.Parameter of Interest The mean time required to complete a federal income tax return Example 2.Confidence Interval Criteria a.Assumptions: Sampled population assumed normal, unknown b.Test statistic: t will be used c.Confidence level: = 0.95

13 13 ES9 3.The Sample Evidence: Solution Continued 5.The Results: 6.62 to 8.98 is the 95% confidence interval for 4.The Confidence Interval a.Confidence coefficients: b.Maximum error: c.Confidence limits: 18.1)5578.0)(12.2( )12.2( n s E t (df, /2) = t (16, 0.025) = 2.12 t (16, 0.025)

14 14 ES9 1.The t-statistic is used to complete a hypothesis test about a population mean Hypothesis-Testing Procedure 2.The test statistic: 3.The calculated t is the number of estimated standard errors is from the hypothesized mean 4.Probability-Value or Classical Approach

15 15 ES9 Example: A random sample of 25 students registering for classes showed the mean waiting time in the registration line was 22.6 minutes and the standard deviation was 8.0 minutes. Is there any evidence to support the student newspapers claim that registration time takes longer than 20 minutes? Use = 0.05 and assume waiting time is approximately normal. Solution: 1.The Set-up a.Population parameter of concern: the mean waiting time spent in the registration line b.State the null and alternative hypotheses: H o : = 20 ( ) (no longer than) H a : > 20 (longer than) Example

16 16 ES9 2.The Hypothesis Test Criteria a.Check the assumptions: The sampled population is approximately normal b.Test statistic: t* with df = n - 1 = 24 c.Level of significance: = 0.05 Solution Continued 3.The Sample Evidence a.Sample information: b.Calculate the value of the test statistic:

17 17 ES9 Using the p-Value Procedure: 4.The Probability Distribution a.The p-value: )24df with,625.1*(P tP Solution Continued Notes: If this hypothesis test is done with the aid of a computer, most likely the computer will compute the p-value for you Using Table 6: place bounds on the p-value Using Table 7: read the p-value directly from the table for many situations: Using Table 6: 0.05 < P < 0.10 Using Table 7: b.The p-value is not smaller than the level of significance, P

18 18 ES9 Using the Classical Procedure: 4.The Probability Distribution a.The critical value: t (24, 0.05) = 1.71 b.t* is not in the critical region 5.The Results a.Decision: Fail to reject H o b.Conclusion:There is insufficient evidence to show the mean waiting time is greater than 20 minutes at the 0.05 level of significance Solution Continued

19 19 ES9 Example: A new study indicates that higher than normal (220) cholesterol levels are a good indicator of possible heart attacks. A random sample of 27 heart attack victims showed a mean cholesterol level of 231 and a standard deviation of 20. Is there any evidence to suggest the mean cholesterol level is higher than normal for heart attack victims? Use = 0.01 Solution: 1.The Set-up a.Population parameter of concern: The mean cholesterol level of heart attack victims b.State the the null and alternative hypothesis: H o : = 220 ( ) (mean is not greater than 220) H a : > 220 (mean is greater than 220) Example

20 20 ES9 2.The Hypothesis Test Criteria a.Assumptions: We will assume cholesterol level is at least approximately normal. b.Test statistic: t* ( unknown), df = n 1 = 26 c.Level of significance: = 0.01 (given) Solution Continued 3.The Sample Evidence a.Sample information: b.Calculate the value of the test statistic:

21 21 ES9 4.The Probability Distribution a.The critical value: t (26, 0.01) = 2.48 Solution Continued 5.The Results a.Decision: Reject H 0 b.Conclusion:At the 0.01 level of significance, there is sufficient evidence to suggest the mean cholesterol level in heart attack victims is higher than normal t (26, 0.01) * b.t* falls in the critical region

22 22 ES9 9.2 ~ Inferences About the Binomial Probability of Success Possibly the most common inference of all Many examples of situations in which we are concerned about something either happening or not happening Two possible outcomes, and multiple independent trials

23 23 ES9 1.p: the binomial parameter, the probability of success on a single trial Background 2. : the observed or sample binomial probability 'p x represents the number of successes that occur in a sample consisting of n trials 3.For the binomial random variable x: 4.The distribution of x is approximately normal if n is larger than 20 and if np and nq are both larger than 5

24 24 ES9 1.a mean equal to p, 2.a standard error equal to, and 3.an approximately normal distribution if n is sufficiently large 'p 'p Sampling Distribution of p': If a sample of size n is randomly selected from a large population with p = P(success), then the sampling distribution of p' has: Sampling Distribution of p' In practice, use of the following guidelines will ensure normality: 1.The sample size is greater than 20 2.The sample consists of less than 10% of the population 3.The products np and nq are both larger than 5

25 25 ES9 The assumptions for inferences about the binomial parameter p: The n random observations forming the sample are selected independently from a population that is not changing during the sampling The Assumptions... Confidence Interval Procedure: The unbiased sample statistic p' is used to estimate the population proportion p The formula for the confidence interval for p is: n qp p n qp z ( p '' ' to '' ' z (

26 26 ES9 Example: A recent survey of 300 randomly selected fourth graders showed 210 participate in at least one organized sport during one calendar year. Find a 95% confidence interval for the proportion of fourth graders who participate in an organized sport during the year. Solution: 1.Describe the population parameter of concern The parameter of interest is the proportion of fourth graders who participate in an organized sport during the year Example 2.Specify the confidence interval criteria a.Check the assumptions The sample was randomly selected Each subjects response was independent

27 27 ES9 b.Identify the probability distribution z is the test statistic p' is approximately normal c.Determine the level of confidence: 0.95 Solution Continued 3.Collect and present sample evidence Sample information: n = 300, and x = 210 The point estimate:

28 28 ES9 4.Determine the confidence interval a.Determine the confidence coefficients: Using Table 4, Appendix B: z ( /2) = z (0.025) = 1.96 Solution Continued c.Find the lower and upper confidence limits: )0265.0)(96.1(0007.0)96.1( 300 )30.0)(70.0( 96.1 '' n qp E b.The maximum error of estimate: z ( /2)

29 29 ES9 d.The Results to is a 95% confidence interval for the true proportion of fourth graders who participate in an organized sport during the year Solution Continued E: maximum error of estimate : confidence level p*: provisional value of p (q* = 1 p*) If no provisional values for p and q are given use p* = q* = 0.5 (Always round up) 2 2 **][ E qp n Sample Size Determination: z ( /2)

30 30 ES9 Example: Determine the sample size necessary to estimate the true proportion of laboratory mice with a certain genetic defect. We would like the estimate to be within with 95% confidence. Example Solution: 1.Level of confidence: = 0.95, z ( /2) = z (0.025) = Desired maximum error is E = No estimate of p given, use p* = q* = Use the formula for n: )015.0( )5.0()5.0()96.1(**][ n E qp n z ( /2)

31 31 ES )015.0( )9875.0()0125.0()96.1(**][ n E qp n Note:Suppose we know the genetic defect occurs in approximately 1 of 80 animals Use p* = 1/80 = 0.125: Note As illustrated here, it is an advantage to have some indication of the value expected for p, especially as p becomes increasingly further from 0.5 z ( /2)

32 32 ES9 Hypothesis-Testing Procedure: For hypothesis tests concerning the binomial parameter p, use the test statistic z*: Example: (Probability-Value Approach) A hospital administrator believes that at least 75% of all adults have a routine physical once every two years. A random sample of 250 adults showed 172 had physicals within the last two years. Is there any evidence to refute the administrator's claim? Use = 0.05 Hypothesis-Testing Procedure

33 33 ES9 Solution 1.The Set-up a.Population parameter of concern: the proportion of adults who have a physical every two years b.State the null and alternative hypotheses: H o : p = 0.75 (>) H a : p < The Hypothesis Test Criteria a.Assumptions: 250 adults independently surveyed b.Test statistic: z* n = 250 np = (250)(0.75) = > 5 nq = (250)(0.25) = 62.5 > 5 c.Level of significance: = 0.05

34 34 ES9 Solution Continued 3.The Sample Evidence a.Sample information: b.The test statistic: 4.The Probability Distribution a.The p-value: Use Table 3 Appendix B, Table 5 Appendix B, or use a computer

35 35 ES )26.2( zPP p-value Solution Continued b.The p-value is smaller than the level of significance, 5.The Results a.Decision: Reject H o b.Conclusion:There is evidence to suggest the proportion of adults who have a routine physical exam every two years is less than 0.75 at the 0.05 level of significance

36 36 ES9 Example:(Classical Procedure) A university bookstore employee in charge of ordering texts believes 65% of all students sell their statistics books back to the bookstore at the end of the class. To test this claim, 200 statistics students are selected at random and 141 plan to sell their texts back to the bookstore. Is there any evidence to suggest the proportion is different from 0.65? Use = 0.01 Solution: 1.The Set-up a.Population parameter of concern: p = the proportion of students who sell their statistics books back to the bookstore b.The null and alternative hypotheses: H o : p = 0.65 H a : Example

37 37 ES9 Solution Continued 2.The Hypothesis Test Criteria a.Assumptions: Sample randomly selected. Each subjects response was independent of other responses. b.Test statistic: z* n = 200 np = (200)(0.65) = 130 > 5 ; nq = (200)(0.35) = 70 > 5 c.Level of significance: = The Sample Evidence a. b.Calculate the value of the test statistic:

38 38 ES9 4.The Probability Distribution a.The critical value: z (0.005) = 2.58 Solution Continued 5.The Results a.Decision: Do not reject H o b.Conclusion:There is no evidence to suggest the true proportion of students who sell their statistics texts back to the bookstore is different from 0.65 at the 0.05 level of significance b.z* is not in the critical region *

39 39 ES9 Notes 1.There is a relationship between confidence intervals and two- tailed hypothesis tests when the level of confidence and the level of significance add up to 1 2.The confidence interval and the width of the noncritical region are the same 3.The point estimate is the center of the confidence interval, and the hypothesized mean is the center of the noncritical region 4.If the hypothesized value of p is contained in the confidence interval, then the test statistic will be in the noncritical region 5.If the hypothesized value of p does not fall within the confidence interval, then the test statistic will be in the critical region

40 40 ES9 9.3 ~ Inferences About Variance & Standard Deviation Problems often arise that require us to make inferences about variability Usually use variance to make inferences about variability Inferences about the variance of a normally distributed population use the chi-square, 2, distribution

41 41 ES9 Background 1.The chi-square distributions are a family of probability distributions Properties of the Chi-Square Distribution: 1. 2 is nonnegative in value; it is zero or positively valued 2.Each distribution is identified by a parameter called the number of degrees of freedom 2. 2 is not symmetrical; it is skewed to the right 3. 2 is distributed so as to form a family of distributions, a separate distribution for each different number of degrees of freedom

42 42 ES9 2 2, df = 6 2, df = 10 2, df = 20 Various Chi-Square Distributions

43 43 ES9 Critical Values for Chi-Square Distribution 1.Table 8 in Appendix B 2. 2 (df, ) The symbol used to identify the critical value of a chi-square distribution with df degrees of freedom It denotes a point on the measurement axis so that there is of the area to the right of that point 3.Since the chi-square distribution is not symmetrical, the critical values associated with right and left tails are given separately in Table 8

44 44 ES (df, ) 2 Distribution Showing 2 (df, )

45 45 ES9 Example: Find the value of 2 (18, 0.025) Example Portion of Table (18, 0.025)

46 46 ES9 0 2 (12, 0.99) Example: Find the value of 2 (12, 0.99) Example Portion of Table

47 47 ES9 Notes 1.When df > 2, the mean value of the chi-square distribution is df 2.The mean is located to the right of the mode (the value where the curve reaches it high point) and just to the right of the median (the value that splits the distribution, 50% on either side) mode x x ~ 0 2 Illustration:

48 48 ES9 The Assumptions... The assumptions for inferences about the variance 2 or standard deviation The sample population is normally distributed Hypothesis-Testing Procedure: 1.The statistical procedures for standard deviation are very sensitive to nonnormal distributions (skewness, in particular). This makes it difficult to decide if a significant result is due to sample evidence or a violation of assumptions. 2.The test statistic: If the random sample is drawn from a normal population with known variance 2, then 2 * has a chi-square distribution with n 1 degrees of freedom )1( * sn

49 49 ES9 Example:A machine used to fill 5-gallon buckets of driveway sealer has a standard deviation of 2.5 ounces. A random sample of 24 buckets showed a standard deviation of 2.9 ounces. Is there any evidence to suggest an increase in variability at the 0.05 level of significance? Assume the amount of driveway sealer in a bucket is normally distributed. Solution: 1.The Set-up a.Population parameter of concern: the variance 2 for the amount of driveway sealer in a 5-gallon bucket b.State the null and alternative hypotheses: H o : 2 = 6.25 ( ) (variance is not larger than 6.25) H a : 2 > 6.25 (variance is larger than 6.25) Example

50 50 ES9 2.The Hypothesis Test Criteria a.Check the assumptions: The sample population is assumed to be normally distributed b.Test statistic: 2 * with df = n 1 = 24 1 = 23 c.Level of significance: = 0.05 Solution Continued 3.The Sample Evidence a.Sample information: n = 24, 2 = (2.9) 2 = 8.41 b.Calculate the value of the test statistic:

51 51 ES9 4.The Probability Distribution (p-Value Approach) a.The p-value: Use Table 8 Appendix B, or use a computer 0.10 < P < 0.25 b.The p-value is larger than the level of significance, Solution Continued 5.The Results a.Decision: Fail to reject H 0 b.Conclusion:There is insufficient evidence to show the variability has increased at the 0.05 level of significance 4.The Probability Distribution (Classical Approach) a.The critical value: 2 (23, 0.05) = 35.2 b. 2 * is not is the critical region ~ or ~

52 52 ES9 Notes 1.Use Table 8 in Appendix B to place bounds on the p-value 2.A calculator or computer may be used to find the exact p-value. Use the 2 probability or the 2 cumulative probability distribution commands. 3.If the sample data is skewed, only one outlier can greatly influence the standard deviation. Therefore, it is very important, especially in small samples, that the sampled population be normal. Otherwise, the statistical procedure is not reliable.


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