BNAD277 Study Guide Test 2
BNAD277 Study Guide Test 2 BNAD277
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Date Created: 03/28/16
BNAD277 Test #2 Study Guide Bnad277: Chapter 8 Notes Hypothesis Testing: - Hypothesis testing can be used to determine whether a statement about the value of a population parameter should or should not be rejected. - The null hypothesis, denoted by H ,0is a tentative assumption about a population parameter. - The alternative hypothesis, denoted by H , as the opposite of what is stated in the null hypothesis. - The hypothesis testing procedure uses data from a sample to test the two competing statements indicated by H an0 H . a - Developing a Null and Alternative Hypotheses: o Many applications of hypothesis testing involve an attempt to gather evidence in support of a research hypothesis. o In such cases, it is often best to begin with the alternative hypothesis and make it the conclusion that the researcher hopes to support. o The conclusion that the research hypothesis is true is made if the sample data provide sufficient evidence to show that the null hypothesis can be rejected. o It is not always obvious how the null and alternative hypotheses should be formulated. o Care must be taken to structure the hypotheses appropriately so that the test conclusion provides the information the researcher wants. o The context of the situation is very important in determining how the hypotheses should be stated. o In some cases it is easier to identify the alternative hypothesis first. In other cases the null is easier. o Correct hypothesis formulation will take practice. o Summary: ▯ The equality part of the hypotheses always appears in the null hypothesis. ▯ In general, a hypothesis test about the value of a population mean m must take one of the following three forms (where m 0s the hypothesized value of the population mean). ▯ - Type 1 Error: o Because hypothesis tests are based on sample data, we must allow for the possibility of errors. o A Type I error is rejecting H when it is true. 0 o The probability of making a Type I error when the null hypothesis is true as an equality is called the level of significance. o Applications of hypothesis testing that only control the Type I error are often called significance tests. - Type 2 Error: o A Type II error is accepting H whe0 it is false. o It is difficult to control for the probability of making a Type II error. o Statisticians avoid the risk of making a Type II error by using “do not reject H ” and not “accept H ”. 0 0 - P value Approach to One-Tailed Hypothesis Testing: o The p-value is the probability, computed using the test statistic,that measures the support (or lack of support) provided by the sample for the null hypothesis. o Reject H if0the p-value < a. - Steps of Hypothesis Testing: o Step 1. Develop the null and alternative hypotheses. o Step 2. Specify the level of significance a. o Step 3. Collect the sample data and compute the value of the test statistic. o Step 4. Use the value of the test statistic to compute the p-value. o Reject H if 0-value < a. - P Value Approach to Two-Tailed Hypothesis Testing: o 1. Compute the value of the test statistic z. o 2. If z is in the upper tail (z > 0), compute the probability that z is greater than or equal to the value of the test statistic. If z is in the lower tail (z < 0), compute the probability that z is less than or equal to the value of the test statistic. o 3. Double the tail area obtained in step 2 to obtain the p –value. o 4. Reject H if 0he p-value < a . - Tests About a Population Mean: o Test Statistic: ▯ This test statistic has a t distribution with n - 1 degrees of freedom. - P -Values and the t Distribution: o The format of the t distribution table provided in most statistics textbooks does not have sufficient detail to determine the exact p- value for a hypothesis test. o However, we can still use the t distribution table to identify a range for the p-value. o An advantage of computer software packages is that the computer output will provide the p-value for the t distribution. - Summary of Forms for Null and Alternative Hypotheses About a Population Proportion: o The equality part of the hypotheses always appears in the null hypothesis. o In general, a hypothesis test about the value of a population proportion pmust take one of the following three forms (where p is 0 the hypothesized value of the population proportion). Bnad277: Chapter 10 Notes Inference About Means and Proportions with 2 Populations: - Inferences About the Difference Between two Population Means: s and s 1 2 Known o Interval Estimation of m – 1 2 o Hypothesis Tests About m – m 1 2 o Let m e1ual the mean of population 1 and m equal 2he mean of population 2. o The difference between the two population means is m - m . 1 2 o To estimate m - 1 , w2 will select a simple random sample of size n from population 1 and a simple random sample of size n from 1 2 population 2. - Sampling Distribution of x bar1- x bar2 o Expected Value: o Standard Deviation (Standard Error): o Interval Estimate: o Example: Par, Inc. is a manufacturer of golf equipment and has developed a new golf ball that has been designed to provide “extra distance.” In a test of driving distance using a mechanical driving device, a sample of Par golf balls was compared with a sample of golf balls made by Rap, Ltd., a competitor. The sample statistics appear on the next slide. ▯ s 1 15 yards and s = 22 yards o Let us develop a 95% confidence interval estimate of the difference between the mean driving distances of the two brands of golf ball. o Point Estimate: 295-278= 17 yards o We are 95% confident that the difference between the mean driving distances of Par, Inc. balls and Rap,Ltd. balls is 11.86 to 22.14 yards. - Inferences About the Difference Between two Population Means: s and 1 s 2nknown o Interval Estimation of m –1m 2 o Hypothesis Tests About m – m 1 2 o When s and s are unknown, we will: 1 2 ▯ use the sample standard deviations s and 1 as es2imates of s 1nd s , 2nd ▯ replace za/2with ta/2 o Interval Estimate: o Where the degrees of freedom are: o Example: Specific Motors of Detroit has developed a new Automobile known as the M car. 24 M cars and 28 J cars (from Japan) were road tested to compare miles-per-gallon (mpg) performance. The sample statistics are shown on the next slide. o Let us develop a 90% confidence interval estimate of the difference between the mpg performances of the two models of automobile. ▯ Point Estimate: 29.8-27.3= 2.5mpg o Hypothesis: o Test Statistic: o - Inferences About the Difference Between two Population Means: Matched Samples o With a matched-sample design each sampled item provides a pair of data values. o This design often leads to a smaller sampling error than the independent-sample design because variation between sampled items is eliminated as a source of sampling error. o Example: A Chicago-based firm has documents that must be quickly distributed to district offices throughout the U.S. The firm must decide between two delivery services, UPX (United Parcel Express) and INTEX (International Express), to transport its documents. In testing the delivery times of the two services, the firm sent two reports to a random sample of its district offices with one report carried by UPX and the other report carried by INTEX. Do the data on the next slide indicate a difference in mean delivery times for the two services? Use a .05 level of significance. ▯ Step 1: Develop the hypotheses: ▯ Step 2: Specify the level of significance: alpha=.05 ▯ Step 3: Computer the value of the test statistic: ▯ Step 4: Compute the p-value: • For t = 2.94 and df = 9, the p–value is between.02 and .01. (This is a two-tailed test, so we double the upper-tail areas of .01 and .005.) ▯ Step 5: Determine whether to reject H : 0 • Because p–value < a = .05, we reject H . 0 • We are at least 95% confident that there is a difference in mean delivery times for the two services. Bnad277: Chapter 12 Notes Tests of Goodness of Fit, Independence, and Multiple Proportions: - In this chapter we introduce three additional hypothesis-testing procedures. 2 - The test statistic and the distribution used are based on the chi-square (c ) distribution. - In all cases, the data are categorical. - Goodness of Fit Test: Multinomial Probability Distribution: o 1. State the null and alternative hypotheses: o 2. Select a random sample and record the observed frequency, f , i for each of the k categories. o 3. Assuming H is t0ue, compute the expected frequency, e , in i each category by multiplying the category probability by the sample size. o 4. Compute the value of the test statistic. o 5. Rejection Rule: o Example: Finger Lakes Homes manufactures four models of prefabricated homes, a two-story colonial, a log cabin, a split-level, and an A-frame. To help in production planning, management would like to determine if previous customer purchases indicate that there is a preference in the style selected. ▯ The number of homes sold of each model for 100 sales over the past two years is shown below. ▯ Hypotheses: ▯ Reject Rule: Reject H 0f p-value </ .05 or x > 7.815 ▯ Conclusion Using the Critical Value Approach: • X = 10 >/ 7.815 • We reject, at the .05 level of significance, the assumption that there is no home-style preference. - Test of Independence: o 1. Set up the null and alternative hypotheses. o 2. Select a random sample and record the observed frequency, f , ij for each cell of the contingency table. o 3. Compute the expected frequency, e , for eijh cell. o 4. Compute the test statistic: o 5. Determine the rejection rule: - Testing the Equality of Population Proportions for Three or More Populations: o Using the notation ▯ p = population proportion for population 1 1 ▯ p2= population proportion for population 2 ▯ and p k population proportion for population k o The hypotheses for the equality of population proportions for k > 3 populations are as follows: o If H 0annot be rejected, we cannot detect a difference among the k population proportions. o If H 0an be rejected, we can conclude that not all k population proportions are equal. o Further analyses can be done to conclude which population proportions are significantly different from others. ▯ Example: Finger Lakes Homes manufactures three models of prefabricated homes, a two-story colonial, a log cabin, and an A-frame. To help in product-line planning, management would like to compare the customer satisfaction with the three home styles. ▯ p = proportion likely to repurchase a Colonial for the 1 population of Colonial owners ▯ p2= proportion likely to repurchase a Log Cabin for the population of Log Cabin owners ▯ p = proportion likely to repurchase an A-Frame for the 3 population of A-Frame owners ▯ Observed Frequencies: ▯ Next, determine the expected frequencies under the assumption H i0 correct. ▯ Expected Frequencies: ▯ Next, compute the value of the chi-square test statistic: ▯ Computation of the Chi-Square Test Statistic: ▯ Rejection Rule: Because c = 8.670 is between 9.210 and 7.378, the area in the upper tail of the distribution is between .01 and .025. ▯ The p-value < a. We can reject the null hypothesis. ▯ We have concluded that the population proportions for the three populations of home owners are not equal. ▯ To identify where the differences between population proportions exist, we will rely on a multiple comparisons procedure. - Multiple Comparisons Procedure: o We begin by computing the three sample proportions: o We will use a multiple comparison procedure known as the Marascuillo procedure. o Marascuillo Procedure: We compute the absolute value of the pair wise difference between sample proportions: o For each pair-wise comparison compute a critical value as follows: o Pair-wise Comparison Tests:
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