Inferential Statistics: Hypothesis Testing College of Engineering and Architecture University of Nueva Caceres
Objective State the null and alternative hypotheses Find critical values for the z test . Test means when s is known, using the z test . Test means when σ is unknown, using the t test . Test variances or standard deviations, using the chi-square test Test hypotheses, using confidence intervals
Introduction Researchers are interested in answering many types of questions . For example, a scientist might want to know whether there is evidence of global warming . A physician might want to know whether a new medication will lower a person’s blood pressure . Automobile manufacturers are interested in determining whether a new type of seat belt will reduce the severity of injuries caused by accidents . These types of questions can be addressed through statistical hypothesis testing, which is a decision-making process for evaluating claims about a population.
Topics Steps in Hypothesis Testing—Traditional Method z-Test for a Mean t - Test for a Mean X 2 Test for a Variance or Standard Deviation
Steps in Hypothesis Testing Traditional Method
Statistical Hypothesis A statistical hypothesis is a conjecture about a population parameter. This conjecture may or may not be true.
Two Types of Statistical Hypothesis Null Hypothesis ( H ) is a statistical hypothesis that states that there is no difference between a parameter and a specific value, or that there is no difference between two parameters . Alternative hypothesis ( H 1 ) is a statistical hypothesis that states the existence of a difference between a parameter and a specific value, or states that there is a difference between two parameters .
Classification Of Alternative Hypothesis Non – directional Hypothesis The one which asserts the value is different from another Makes use of the “not equal to” sign two – sided hypothesis
Classification Of Alternative Hypothesis Directional Hypothesis An assertion that one measure is less than or greater than another measure of similar nature Involves one of the order relatives, “less than” or “greater than” One – sided hypothesis
How Hypotheses Should Be S tated?
Situation A A medical researcher is interested in finding out whether a new medication will have any undesirable side effects. The researcher is particularly concerned with the pulse rate of the patients who take the medication. Will the pulse rate increase, decrease , or remain unchanged after a patient takes the medication? Since the researcher knows that the mean pulse rate for the population under study is 82 beats per minute
Situation A the hypotheses for this situation are Null hypothesis : the mean will remain unchanged Alternative hypothesis : the mean will be different This test is called a two-tailed test since the possible side effects of the medicine could be to raise or lower the pulse rate.
Situation B A chemist invents an additive to increase the life of an automobile battery . If the mean lifetime of the automobile battery without the additive is 36 months, then her hypotheses are : This test is called right-tailed , since the interest is in an increase only
Situation C A contractor wishes to lower heating bills by using a special type of insulation in houses. If the average of the monthly heating bills is $78, 78 This test is a left-tailed test , since the contractor is interested only in lowering heating costs
Conclusions in Hypothesis Testing reject H in favor of H 1 because of sufficient evidence in the data fail to reject H because of insufficient evidence in the data .
Illustration: H : defendant is innocent , H 1 : defendant is guilty .
Possible Outcomes of a Hypothesis Test A type I error occurs if you reject the null hypothesis when it is true. A type II error occurs if you do not reject the null hypothesis when it is false.
Level of Significance is the maximum probability of committing a type I error. This probability is symbolized by a (Greek letter alpha). That is, P(type I error) = α . levels: 0.10, 0.05 , and 0.01 Example: when α = 0.05, there is a 5% chance of rejecting a true null hypothesis
The critical or rejection region is the range of test values that indicates that there is a significant difference and that the null hypothesis should be rejected. The noncritical or nonrejection region is the range of test values that indicates that the difference was probably due to chance and that the null hypothesis should not be rejected. The critical value separates the critical region from the noncritical region. The symbol for critical value is C.V.
one-tailed test indicates that the null hypothesis should be rejected when the test value is in the critical region on one side of the mean. A one-tailed test is either a right-tailed test or left-tailed test.
Right Tailed Test Situation B A chemist invents an additive to increase the life of an automobile battery. If the mean lifetime of the automobile battery without the additive is 36 months, then her hypotheses are :
Left-Tailed Test Situation C A contractor wishes to lower heating bills by using a special type of insulation in houses. If the average of the monthly heating bills is $78, 78
Two-Tailed Test the null hypothesis should be rejected when the test value is in either of the two critical regions . the critical region must be split into two equal parts. If α = 0.01, then one-half of the area, or 0.005, must be to the right of the mean and one-half must be to the left of the mean
Two-Tailed Test Situation A A medical researcher is interested in finding out whether a new medication will have any undesirable side effects. The researcher is particularly concerned with the pulse rate of the patients who take the medication. Will the pulse rate increase, decrease, or remain unchanged after a patient takes the medication?
Tw0-Tailed Test
Sample Problem: A manufacturer of a certain brand of rice cereal claims that the average saturated fat content does not exceed 1.5 grams per serving. State the null and alternative hypotheses to be used in testing this claim and determine where the critical region is located.
Sample Problem: A real estate agent claims that 60% of all private residences being built today are 3-bedroom homes. To test this claim, a large sample of new residences is inspected; the proportion of these homes with 3 bedrooms is recorded and used as the test statistic. State the null and alternative hypotheses to be used in this test and determine the location of the critical region.
Solving Hypothesis-Testing Problems (Traditional Method) 1. State the null and alternative hypotheses. 2. Choose a fixed significance level α . 3. Choose an appropriate test statistic and establish the critical region based on α . 4. Reject H 0 if the computed test statistic is in the critical region. Otherwise, do not reject. 5. Draw scientific or engineering conclusions.
Summarizing Results
Single Sample: Tests Concerning a Single Mean
Tests on a Single Mean (Variance Known) is a statistical test for the mean of a population. It can be used either when n 30 or when the population is normally distributed and σ is known.
EXAMPLE A manufacturer of sports equipment has developed a new synthetic fishing line that the company claims has a mean breaking strength of 8 kilograms with a standard deviation of 0.5 kilogram. Test the hypothesis that μ = 8 kilograms against the alternative that μ = 8 kilograms if a random sample of 50 lines is tested and found to have a mean breaking strength of 7.8 kilograms. Use a 0.01 level of significance.
EXAMPLE A random sample of 100 recorded deaths in the United States during the past year showed an average life span of 71.8 years. Assuming a population standard deviation of 8.9 years, does this seem to indicate that the mean life span today is greater than 70 years? Use a 0.05 level of significance.
EXAMPLE An electrical firm manufactures light bulbs that have a lifetime that is approximately normally distributed with a mean of 800 hours and a standard deviation of 40 hours. Test the hypothesis that μ = 800 hours against the alternative, μ 800 hours, if a random sample of 30 bulbs has an average life of 788 hours. Use 0.05 level of significance.
Tests on a Single Sample (Variance Unknown) is a statistical test for the mean of a population and is used when the population is normally or approximately normally distributed and σ is unknown
Tests on a Single Sample (Variance Unknown) is a statistical test for the mean of a population and is used when the population is normally or approximately normally distributed and σ is unknown As the degrees of freedom get larger, the critical values approach the z values
Zee vs. Tee z - test t - test σ – known σ - unknown n > 30 Level of significance Degree of freedom v = n – 1 z - test t - test σ – known σ - unknown n > 30 Level of significance Degree of freedom v = n – 1
EXAMPLE The Edison Electric Institute has published figures on the number of kilowatt hours used annually by various home appliances. It is claimed that a vacuum cleaner uses an average of 46 kilowatt hours per year. If a random sample of 12 homes included in a planned study indicates that vacuum cleaners use an average of 42 kilowatt hours per year with a sample standard deviation of 11.9 kilowatt hours, does this suggest at the 0.05 level of significance that vacuum cleaners use, on average, less than 46 kilowatt hours annually? Assume the population of kilowatt hours to be normal.
EXAMPLE A corporation sets its annual budget for a new plant on the assumption that the average weekly cost for repairs is to be $1,200. To see whether this claim is realistic, 10 weekly repairs cost figures are obtained from similar plants. The sample is assumed to be random and yields a mean of $1,290 and a standard deviation of $110. Does this sample indicate that $1,200 is not good assumed value for the mean weekly cost of repairs? Use a 0.05 level of significance. Assume normality of weekly repair costs.
EXAMPLE A report from 6 years ago indicated that the average gross salary for a business analyst was $ 69,873. Since this survey is now out – dated, the Bureau of Labor Statistics wishes to test this figure against current salaries to see if the current salaries are statistically different from the old ones. Based on the samples, it is found that the mean is $ 79, 180 and the standard deviation is $14, 985. for this study, the bureau will take a sample of 12 current salaries. Use a 0.05 level of significance.
Test on the Variance and Standard Deviation of a Normal Distribution
Testing for Variance : Chi – Square Test A chi – squared test, also written as x 2 is any statistical hypothesis test wherein the sampling distribution of the test statistic is a chi – squared distribution when the null hypothesis is true. Test for normality
Properties of Chi – Square test for Variance and Standard deviation 1. All chi-square values are greater than or equal to 0. 2. The chi-square distribution is a family of curves based on the degrees of freedom. 3. The area under each chi-square distribution is equal to 1. 4. The chi-square distributions are positively skewed.
Chi – Square Test used to test a claim about a single variance or standard deviation. σ² s ²
Test for normality NULL HYPOTHESIS σ ² = σ ALTERNATIVE HYPOTHESIS σ ² < σ σ ² > σ σ ² σ
EXAMPLE A manufacturer of car batteries claims that the life of the company’s batteries is approximately normally distributed with a standard deviation equal to 0.9 year . If a random sample of 10 of these batteries has a standard deviation of 1.2 years, do you think that σ > . 9 year? Use a 0.05 level of significance.
EXAMPLE The standard deviation for the Math SAT test is 100. The variance is 10,000. An instructor wishes to see if the variance of the 23 randomly selected students in her school is less than 10,000. The variance for the 23 test scores is 7225. Is there enough evidence to support the claim that the variance of the students in her school is less than 10,000 at α = 0.05 ? Assume that the scores are normally distributed.
EXAMPLE Past experience indicates that the time required for high school seniors to complete a standardized test is a normal random variable with a standard deviation of 6 minutes. Test the hypothesis that σ = 6 against the alternative that σ < 6 if a random sample of the test times of 20 high school seniors has a standard deviation s = 4 . 51. Use a 0.05 level of significance.
Test on a Single Proportion
Many hypothesis-testing situations involve proportions. A proportion is the same as a percentage of the population . • 59% of consumers purchase gifts for their fathers. • 85% of people over 21 said they have entered a sweepstakes . • 51% of Americans buy generic products . • 35% of Americans go out for dinner once a week .
A hypothesis test involving a population proportion can be considered as a binomial experiment when there are only two outcomes and the probability of a success does not change from trial to trial. Recall that the mean is µ = np and the standard deviation is for the binomial distribution.
Hypotheses: p < p p>p p ≠ p
Testing a Proportion
EXAMPLE The Gallup Crime Survey stated that 23% of gun owners are women. A researcher believes that in the area where he lives, the percentage is less than 23%. He randomly selects a sample of 100 gun owners and finds that 11% of the gun owners are women. At a 0.01, is the percentage of female gun owners in his area less than 23%.
EXAMPLE An attorney claims that more than 25% of all lawyers advertise. A random sample of 200 lawyers in a certain city showed that 63 had used some form of advertising. At a 0.05, is there enough evidence to support the attorney’s claim ?
EXAMPLE A builder claims that heat pumps are installed in 70% of all homes being constructed today in the city of Richmond, Virginia. Would you agree with this claim if a random survey of new homes in this city showed that 8 out of 15 had heat pumps installed? Use a 0.10 level of significance.
EXAMPLE A new radar device is being considered for a certain missile defense system. The system is checked by experimenting with aircraft in which a kill or a no kill is simulated. If, in 300 trials, 250 kills occur, accept or reject, at the 0.04 level of significance, the claim that the probability of a kill with the new system does not exceed the 0.8 probability of the existing device.
EXAMPLE A commonly prescribed drug for relieving nervous tension is believed to be only 60 % effective. Experimental results with a new drug administered to a random sample of 100 adults who were suffering from nervous tension show that 70 received relief . Is this sufficient evidence to conclude that the new drug is superior to the one commonly prescribed? Use a 0.05 level of significance.