MTH3003 PJJ SEM II 2014/2015 F2F II 12/4/2015

 ASSIGNMENT :25% Assignment 1 (10%) Assignment 2 (15%)  Mid exam :30% Part A (Objective) Part B (Subjective)  Final Exam: 40% Part A (Objective) Part B (Subjective - Short) Part C (Subjective – Long)

CHAPTER 8 ESTIMATION (LARGE SAMPLE) o Definition o Types of estimators:  Point estimator  Interval estimator

Key Concepts I. Types of Estimators 1. Point estimator: a single number is calculated to estimate the population parameter. Interval estimator 2. Interval estimator: two numbers are calculated to form an interval that contains the parameter. II. Properties of Good Estimators 1. Unbiased: the average value of the estimator equals the parameter to be estimated. 2. Minimum variance: of all the unbiased estimators, the best estimator has a sampling distribution with the smallest standard error. 3. The margin of error measures the maximum distance between the estimator and the true value of the parameter.

The Margin of Error  Margin of error:  Margin of error: The maximum error of estimation, is the maximum likely difference observed between sample mean x and true population mean µ, calculated as :

Key Concepts III. Large-Sample Point Estimators To estimate one of four population parameters when the sample sizes are large, use the following point estimators with the appropriate margins of error.

Example 1 A homeowner randomly samples 64 homes similar to her own and finds that the average selling price is $252,000 with a standard deviation of $15,000. Estimate the average selling price for all similar homes in the city.

A quality control technician wants to estimate the proportion of soda cans that are underfilled. He randomly samples 200 cans of soda and finds 10 underfilled cans. Example 2

Key Concepts IV. Large-Sample Interval Estimators To estimate one of four population parameters when the sample sizes are large, use the following interval estimators.

Example 3 A random sample of n = 50 males showed a mean average daily intake of dairy products equal to 756 grams with a standard deviation of 35 grams. Find a 95% confidence interval for the population average . 1.96

Of a random sample of n = 150 college students, 104 of the students said that they had played on a soccer team during their K-12 years. Estimate the proportion of college students who played soccer in their youth with a 98% confidence interval. Example

Example 5 Compare the average daily intake of dairy products of men and women using a 95% confidence interval. Avg Daily IntakesMenWomen Sample size50 Sample mean Sample Std Dev

Example 6 Compare the proportion of male and female college students who said that they had played on a soccer team during their K-12 years using a 99% confidence interval. Youth SoccerMaleFemale Sample size8070 Played soccer

CHAPTER 9 LARGE SAMPLE TESTS OF HYPOTHESES PART I Testing the single mean & single proportion PART II Testing the difference between two means & difference between two proportions

Key Concepts I.Parts of a Statistical Test 1.Null hypothesis: a contradiction of the alternative hypothesis 2.Alternative hypothesis: the hypothesis the researcher wants to support. 3.Test statistic and its p-value: sample evidence calculated from sample data. 4.Rejection region—critical values and significance levels: values that separate rejection and nonrejection of the null hypothesis 5.Conclusion: Reject or do not reject the null hypothesis, stating the practical significance of your conclusion.

Key Concepts II.Errors and Statistical Significance 1.The significance level  is the probability if rejecting H 0 when it is in fact true. 2.The p-value is the probability of observing a test statistic as extreme as or more than the one observed; also, the smallest value of  for which H 0 can be rejected. 3.When the p-value is less than the significance level , the null hypothesis is rejected. This happens when the test statistic exceeds the critical value. 4.In a Type II error,  is the probability of accepting H 0 when it is in fact false. The power of the test is (1   ), the probability of rejecting H 0 when it is false.

Key Concepts III.Large-Sample Test Statistics Using the z Distribution To test one of the four population parameters when the sample sizes are large, use the following test statistics:

Example 1 (testing the single mean) The daily yield for a chemical plant has averaged 880 tons for several years. The quality control manager wants to know if this average has changed. She randomly selects 50 days and records an average yield of 871 tons with a standard deviation of 21 tons.

Using critical value approach: What is the critical value of z that cuts off exactly  /2 =.01/2 =.005 in the tail of the z distribution? Rejection Region: Reject H 0 if z > 2.58 or z < If the test statistic falls in the rejection region, its p-value will be less than a =.01. For our example, z = falls in the rejection region and H 0 is rejected at the 1% significance level.

This is an unlikely occurrence, which happens about 2 times in 1000, assuming  = 880! Using p-value approach:

Example 2 (testing the single proportion) Regardless of age, about 20% of American adults participate in fitness activities at least twice a week. A random sample of 100 adults over 40 years old found 15 who exercised at least twice a week. Is this evidence of a decline in participation after age 40? Use  =.05.

Using critical value approach: What is the critical value of z that cuts off exactly  =.05 in the left-tail of the z distribution? Rejection Region: Reject H 0 if z < If the test statistic falls in the rejection region, its p-value will be less than  =.05. For our example, z = does not fall in the rejection region and H 0 is not rejected. There is not enough evidence to indicate that p is less than.2 for people over 40.

Example 3 (testing difference between two means) Is there a difference in the average daily intakes of dairy products for men versus women? Use a =.05. Avg Daily IntakesMenWomen Sample size50 Sample mean Sample Std Dev3530

Using p-value approach: Since the p-value is greater than  =.05, H 0 is not rejected. There is insufficient evidence to indicate that men and women have different average daily intakes.

Example 4 (testing difference between two proportions) Compare the proportion of male and female college students who said that they had played on a soccer team during their K-12 years using a test of hypothesis. Youth SoccerMaleFemale Sample size8070 Played soccer6539

Using p-value approach: Youth SoccerMaleFemale Sample size8070 Played soccer6539 Since the p-value is less than  =.01, H 0 is rejected. The results are highly significant. There is evidence to indicate that the rates of participation are different for boys and girls.

CHAPTER 10 INFERENCE FROM SMALL SAMPLE PART I Testing the single mean & difference between two means PART II Testing the single variance & ratio of two variances

Key Concepts I. Experimental Designs for Small Samples 1. Single random sample: The sampled population must be normal. 2.Two independent random samples: Both sampled populations must be normal. a. Populations have a common variance   2. b. Populations have different variances 3.Paired-difference or matched-pairs design: The samples are not independent.

Key Concepts II. Statistical Tests of Significance 1.Based on the t, F, and  2 distributions 2.Use the same procedure as in Chapter 9 3. Rejection region  —  critical values and significance levels: based on the t, F, and  2 distributions with the appropriate degrees of freedom 4.Tests of population parameters: a single mean, the difference between two means, a single variance, and the ratio of two variances III. Small Sample Test Statistics To test one of the population parameters when the sample sizes are small, use the following test statistics:

Key Concepts

Testing the single mean  The basic procedures are the same as those used for large samples. For a test of hypothesis: One-tailed (lower-tail)One-tailed (upper-tail)Two-tailed Using p-values or a rejection region based on t distribution with df = n-1

For a 100(1  )% confidence interval for the population mean  Confidence Interval

A sprinkler (sprayer) system is designed so that the average time for the sprinklers to activate after being turned on is no more than 15 seconds. A test of 5 systems gave the following times: 17, 31, 12, 17, 13, 25 Is the system working as specified? Test using  =.05. Example 1

Solution Data: Data: 17, 31, 12, 17, 13, 25 First, calculate the sample mean and standard deviation, using your calculator or the formulas in Chapter 2.

Data: Data: 17, 31, 12, 17, 13, 25 Calculate the test statistic and find the rejection region for  =.05. Rejection Region: Reject H 0 if t > If the test statistic falls in the rejection region, its p-value will be less than  =.05.

Data: Data: 17, 31, 12, 17, 13, 25 Compare the observed test statistic to the rejection region, and draw conclusions. Conclusion: For our example, t = 1.38 does not fall in the rejection region and H 0 is not rejected. There is insufficient evidence to indicate that the average activation time is greater than 15.

Testing the difference between two means (Independent Samples) As in Chapter 9, independent random samples of size n 1 and n 2 are drawn from population 1 and population 2 with means  1 dan  2,and variances and. Since the sample sizes are small, the two populations must be normal  The basic procedures are the same as those used for large samples. For a test of hypothesis: One-tailed (lower-tail)One-tailed (upper-tail) Two-tailed

Interval Estimate of  1 -  2 : Small-Sample Case (n 1 < 30 and/or n 2 < 30) Interval Estimate with  2 Unknown

Instead of estimating each population variance separately, we estimate the common variance with has a t distribution with n 1 +n 2 -2 degrees of freedom. And the resulting test statistic, Test Statistics ( )

How to check the reasonable equality of variance assumption? Test Statistics (cont’d) Rule of Thumb Assume that the variance are equal Do Not Assume that the variance are equal

If the population variances cannot be assumed equal, the test statistic is It has an approximate t distribution with degrees of freedom of Test Statistics ()

Confidence Interval () You can also create a 100(1 -  )% confidence interval for  1 -  2. Remember the three assumptions: 1.Original populations normal 2.Samples random and independent 3.Equal population variances. Remember the three assumptions: 1.Original populations normal 2.Samples random and independent 3.Equal population variances.

Example 2 Two training procedures are compared by measuring the time that it takes trainees to assemble a device. A different group of trainees are taught using each method. Is there a difference in the two methods? Use  = Time to AssembleMethod 1Method 2 Sample size1012 Sample mean3531 Sample Std Dev4.94.5

Solution Hypothesis Equality of Variances Checking Test Statistics

Using critical value approach: What is the critical value of t that cuts off exactly  /2 =.01/2 =.005 in the tail of the t distribution? Critical value: Rejection Region: Reject H 0 if t > or t < If the test statistic falls in the rejection region, its p-value will be less than  =.01 For our example, t = 1.99 falls in the rejection region and H 0 is rejected at the 1% significance level.

The Paired-Difference Test (dependent samples) matched-pairspaired-difference testSometimes the assumption of independent samples is intentionally violated, resulting in a matched-pairs or paired-difference test. By designing the experiment in this way, we can eliminate unwanted variability in the experiment by analyzing only the differences, d i = x 1i – x 2i to see if there is a difference in the two population means,  1 -  2.

Example 3 One Type A and one Type B tire are randomly assigned to each of the rear wheels of five cars. Compare the average tire wear for types A and B using a test of hypothesis. Car12345 Type A Type B But the samples are not independent. The pairs of responses are linked because measurements are taken on the same car.

The Paired-DifferenceTest

Car12345 Type A Type B Difference Solution

Car12345 Type A Type B Difference Rejection region: Reject H 0 if |t| > Conclusion: Since t table = 12.8, H 0 is rejected. There is a difference in the average tire wear for the two types of tires. Solution

Confidence interval (dependent samples) You can construct a 100(1-  )% confidence interval for a paired experiment using

Testing a single variance

A cement manufacturer claims that his cement has a compressive strength with a standard deviation of 10 kg/cm 2 or less. A sample of n = 10 measurements produced a mean and standard deviation of 312 and 13.96, respectively. Do these data produce sufficient evidence to reject the manufacturer’s claim? Use  =.05 A test of hypothesis: H 0 :  2 = 100 (claim is correct) H 1 :  2 > 100 (claim is wrong) A test of hypothesis: H 0 :  2 = 100 (claim is correct) H 1 :  2 > 100 (claim is wrong) uses the test statistic: Example 4

Rejection region: Reject H 0 if     05 . Conclusion: Since   = 17.55, H 0 is rejected. The standard deviation of the cement strengths is more than 10.

Testing the ratio of two variances

An experimenter has performed a lab experiment using two groups of rats. He wants to test H 0 :  1 =  2, but first he wants to make sure that the population variances are equal. Standard (2)Experimental (1) Sample size1011 Sample mean Sample Std Dev Example 5

Standard (2)Experimental (1) Sample size1011 Sample Std Dev We designate the sample with the larger standard deviation as sample 1, to force the test statistic into the upper tail of the F distribution. Solution

The rejection region is two-tailed, with  =.05, but we only need to find the upper critical value, which has  =.025 to its right. From Table 6, with df 1 =10 and df 2 = 9, we reject H 0 if F > CONCLUSION: Reject H 0. There is sufficient evidence to indicate that the variances are unequal. Do not rely on the assumption of equal variances for your t test! The rejection region is two-tailed, with  =.05, but we only need to find the upper critical value, which has  =.025 to its right. From Table 6, with df 1 =10 and df 2 = 9, we reject H 0 if F > CONCLUSION: Reject H 0. There is sufficient evidence to indicate that the variances are unequal. Do not rely on the assumption of equal variances for your t test! Solution