Inferential Statistics Module 7 PDF

Statistical Analysis with Computer Application MODULE 7 INFERENTIAL STATISTICS Dr. Marjeric L. Buenafe Learning Outcomes 01 Understand the deﬁnitions used in hypothesis testing 02 State the null and alternative hypotheses. 03 Find critical values for the z test 04 State the ﬁve steps used in hypothesis testing Learning Outcomes Test means when sigma is known, using the z 05 test. 06 Test means when sigma is unknown, using the t test. LESSONS IN THIS MODULE (Wk 13: 16 - 21 Nov) 7.1. 7.3. Steps in Hypothesis Testing – Traditional t Test for a Mean Method 7.2. z Test for a Mean LESSON 7.1. Steps in Hypothesis Testing – Traditional Method Procedure based on sample evidence and probability theory to determine whether the hypothesis is a reasonable statement. Inferential Statistics Recall the Standard Normal Curve, Z score, area of the curve (P)  0.01  0.01 = = 0.005 = = 0.005 2 2 2 2 p-value H0 Rejection H0 Rejection region region -2.58 -1.55 1.55 2.58 Inferential Statistics There are three methods used to test hypotheses. These are: 1.The traditional method 2.The P-value method 3.The confidence interval method Inferential Statistics A statistical hypothesis is a conjecture about a population parameter. This conjecture may or may not be true. Inferential Statistics Null hypothesis (HO or H0)  a statement about the value of a population parameter (Lind, A., et.al.)  hope to reject  regardless how the problem is stated H0 will always contain the equal sign  point of the testing process, it serves as our working hypothesis  no difference, no effect, no relationship Inferential Statistics Alternative hypothesis (HA or H1)  a statement that is accepted if the sample data provide sufficient evidence that the HO is false  predictive hypothesis  existence difference, effect, relationship  one group is better than the other, non-directional Inferential Statistics Three different statistical studies – Sample of Hypothesis 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, the hypotheses for this situation are H0: μ = 82 and HA: μ ≠ 82 Inferential Statistics Three different statistical studies – Sample of Hypothesis The null hypothesis specifies that the mean will remain unchanged, and the alternative hypothesis states that it will be different. This test is called a two-tailed test (a term that will be formally defined later in this section), since the possible side effects of the medicine could be to raise or lower the pulse rate. Inferential Statistics Three different statistical studies – Sample of Hypothesis 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 H0: μ = 36 and HA: μ > 36 In this situation, the chemist is interested only in increasing the lifetime of the batteries, so her alternative hypothesis is that the mean is greater than 36 months. The null hypothesis is that the mean is equal to 36 months. This test is called right-tailed, since the interest is in an increase only. Inferential Statistics Three different statistical studies – Sample of Hypothesis 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, her hypotheses about heating costs with the use of insulation are H0: μ = $78 and HA: μ < $78 This test is a left-tailed test, since the contractor is interested only in lowering heating costs. Inferential Statistics Two-Tailed Test: non-directional Situation A. H0: μ = 82 and HA: μ ≠ 82 H0 Rejection H0 Rejection region region Inferential Statistics Right-Tailed Test: directional Situation B. H0: μ = 36 and HA: μ > 36 H0 Rejection region Inferential Statistics Left-Tailed Test: directional Situation C. H0: μ = $78 and HA: μ < $78 H0 Rejection region Inferential Statistics A claim, however, can be stated as either the null hypothesis or the alternative hypothesis; but the statistical evidence can only support the claim if it is the alternative hypothesis. Inferential Statistics Table 7.1.1. Common phrases used in hypotheses and conjectures, and the corresponding Symbols. Table 7.1.1. Hypothesis-Testing Common Phrases > < Is greater than Is less than Is above Is below Is higher than Is lower than Is longer than Is shorter than Is bigger than Is smaller than Is increased Is decreased or reduced from Inferential Statistics Table 7.1.1. Common phrases used in hypotheses and conjectures, and the corresponding Symbols. Table 7.1.1. Hypothesis-Testing Common Phrases =  Is equal to Is not equal to Is no different from Is different from Has not changed from Has changed from Is the same as Is not the same as Inferential Statistics Example 6.1.1. State the null and alternative hypotheses for each conjuncture a. A researcher studies gambling in young people. She thinks those who gamble spend more than $30 per day. Solution: H0: μ = $30 and H1: μ > $30 b. A researcher wishes to see if police officers whose spouses work in law enforcement have a lower score on a work stress questionnaire than the average score of 120. Solution: H0: = 120 and HA: μ < 120 Inferential Statistics Example 6.1.1. State the null and alternative hypotheses for each conjuncture c. A teacher feels that if an online textbook is used for a course instead of a hardback book, it may change the students’ scores on a final exam. In the past, the average final exam score for the students was 83. Solution: H0: = 83 and HA: μ ≠ 83 Inferential Statistics The four possibilities are as follows: 1. We reject the null hypothesis when it is true. This would be an incorrect decision and would result in a type I error. 2. We reject the null hypothesis when it is false. This would be a correct decision. 3. We do not reject the null hypothesis when it is true. This would be a correct decision. 4. We do not reject the null hypothesis when it is false. This would be an incorrect decision and would result in a type II error. Inferential Statistics In reality, the null hypothesis may or may not be true, and a decision is made to reject or failed to reject it on the basis of the data obtained from a sample. The four possible outcomes are shown in Figure 6.1.1. Decision Actual Condition Ho is True Ho is False Reject Ho Type I (a) error Correct decision Do not Correct decision Type II (b) error Reject Ho Figure 7.1.1. Four Possible Outcomes (Type I and Type II Errors) Inferential Statistics Setting up Level of Significance The level of significance is the maximum probability of committing a type I error. This probability is symbolized by α (Greek letter alpha). That is, P(type I error) = α. That is, when there is a large difference between the mean obtained from the sample and the hypothesized mean, the null hypothesis is probably not true. Statisticians generally agree on using three arbitrary significance levels: the 0.10, 0.05, and 0.01 levels. That is, if the null hypothesis is rejected, the probability of a type I error will be 10, 5, or 1%, depending on which level of significance is used. Inferential Statistics Test statistic. A value, determined from sample information, used to determine whether to reject the null hypothesis. Inferential Statistics Selecting the Critical Value After a significance level is chosen, a critical value is selected from a table for the appropriate test. The critical value determines the critical and noncritical regions. Standard Normal Cumulative Probability Table (https://www.math.arizona.edu/~jwatkins/normal-table.pdf). Inferential Statistics To obtain the critical value, the researcher must choose an alpha level Inferential Statistics To obtain the critical value, the researcher must choose an alpha level Figure 7.1.3. Two-Tailed Test Figure 7.1.2 (a) Right-Tailed Figure 7.1.2 (b) Left-Tailed Inferential Statistics Summary of Critical Values Figure 7.1.4. One-tailed Test (Left-tailed) Souce: Bluman, 2018 Inferential Statistics Summary of Critical Values Figure 7.1.5. One-tailed Test (Right-tailed) Souce: Bluman, 2018 Inferential Statistics Summary of Critical Values Figure 7.1.5. Two Tailed Test Source: Bluman, 2018 Inferential Statistics Example 7.1.2. Using Standard Normal Cumulative Probability Table, find the critical value(s) for each situation and draw the appropriate figure, showing the critical region. (a) A left-tailed test with α = 0.10. Solution: Step 1. Draw the figure and indicate the appropriate area. Since this is a left-tailed test, the area of 0.10 is located in the left tail, as shown in Figure 7.1.6. Step 2. Find the area closest to 0.1000 in Standard Normal Cumulative Probability Table (https://www.math.arizona.edu/~jwatkins/norma Figure 7.1.6. Critical Value and Critical Region l-table.pdf).. In this case, it is 0.1003. Find the z value that corresponds to the area 0.1003. It is −1.28. See 7.1.6. Inferential Statistics Example 7.1.2. Using Standard Normal Cumulative Probability Table, find the critical value(s) for each situation and draw the appropriate figure, showing the critical region. (b) A two-tailed test with α = 0.02. Solution: Step 1. Draw the figure and indicate the appropriate area. In this case, there are two areas equivalent to α∕2, or 0.02∕2 = 0.01. Step 2. For the left z critical value, find the area closest to α∕2, or 0.02∕2 = 0.01. In this case, it is 0.0099. Figure 7.1.6. Critical Value and For the right z critical value, find the area closest to 1 Critical Region − α∕2, or 1 − 0.02∕2 = 0.9900. In this case, it is 0.9901. Inferential Statistics Example 7.1.2. Using Standard Normal Cumulative Probability Table, find the critical value(s) for each situation and draw the appropriate figure, showing the critical region. (b) A two-tailed test with α = 0.02. Step 2. Find the z values for each of the areas. For 0.0099, z = −2.33. For the area of 0.9901, z = +2.33. See Figure 6.1.6. Figure 7.1.7. Critical Value and Critical Region Inferential Statistics Example 7.1.2. Using Standard Normal Cumulative Probability Table, find the critical value(s) for each situation and draw the appropriate figure, showing the critical region. (c). A right-tailed test with α = 0.005 Solution: Step 1. Draw the figure and indicate the appropriate area. Since this is a right-tailed test, the area 0.005 is located in the right tail, as shown in Figure 7.1.7. Step 2. Find the area closest to 1 − α, or 1 − 0.005 = 0.9950. In this case, it is 0.9949 or 0.9951. Figure 6.1.7. Critical Value and Critical Region Inferential Statistics The two z values corresponding to areas 0.9949 and 0.9951 are +2.57 and +2.58 respectively. Since 0.9500 is halfway between these two z values, find the average of the two z values (+2.57 + 2.58)  2 = +2.575. Figure 6.1.7. Critical Value However, 2.58 is most often used. and Critical Region See Figure 6.1.7. Inferential Statistics For the purpose of this module we will use the simplified Five-Steps Procedure for Testing a Hypothesis (See Figure 6.1.8.) by Lind, Marchal, & Wathen, (2018) Figure 6.1.8. Five-Steps Procedure for Testing a Hypothesis Source: Lind, Marchal, & Wathen, (2018) Inferential Statistics In concluding the results of a statistical study the following table will serve as a guideline of it see Table (Bluman, 2018). Decision Claim Claim is H0 Claim is H1 Reject H0 There is enough There is enough evidence to reject evidence to support the claim… the claim… Do not There is not enough There is not enough Reject H0 evidence to reject evidence to support the claim… the claim… Inferential Statistics Activity 7.1.1. 1. Define null and alternative hypotheses, and give an example of each. 2. What is meant by a type I error? A type II error? How are they related? 3. What is meant by a statistical test? 4. Explain the difference between a one-tailed and a two-tailed test. 5. What is meant by the critical region? The noncritical region? 6. What symbols are used to represent the null hypothesis and the alternative hypothesis? 7. What symbols are used to represent the probabilities of type I and type II errors? Inferential Statistics Activity 7.1.1. 8. Explain what is meant by a significant difference. 9. When should a one-tailed test be used? A two-tailed test? 10. In hypothesis testing, why can’t the hypothesis be proved true? 11. Using the z table (Table E), find the critical value (or values) for each. a. α = 0.10, two-tailed test b. α = 0.01, right-tailed test c. α = 0.005, left-tailed test d. α = 0.01, left-tailed test e. α = 0.05, right-tailed test Inferential Statistics Activity 7.1.1. 12. For each conjecture, state the null and alternative hypotheses. a. The average weight of dogs is 15.6 pounds. b. The average distance a person lives away from a toxic waste site is greater than 10.8 miles. c. The average farm size in 1970 was less than 390 acres. d. The average number of miles a vehicle is driven per year is 12,603. e. The average amount of money a person keeps in his or her checking account is less than $24. LESSON 7.2. z Test for a Mean The observed value is the statistic (such as the sample mean) that is computed from the sample data. The expected value is the parameter (such as the population mean) that you would expect to obtain if the null hypothesis were true—in other words, the hypothesized value. The denominator is the standard error of the statistic being tested (in this case, the standard error of the mean). z Test for a Mean Recall: Five-Step Procedure for Testing a Hypothesis. Step 1. State null and alternate hypotheses Step 2. Select a level of significance Step 3. Identify the test statistic Step 4. Formulate a decision rule Step 5. Take a sample, arrive at decision z Test for a Mean ഥ to Population Mean 𝝁 z Test for Comparing a Sample Means 𝒙 The z Test is a statistical test for the mean of a population. Iy canncncan be used either when n ≥ 30 when the population is normally distributed and σ is known. The formula for the z test is Where 𝑥ҧ = sample mean 𝑥ҧ − 𝜇 𝑥ҧ − 𝜇 𝑧=𝜎 μ = hypothesized population mean or 𝑧= ∙ 𝑛 σ = population standard deviation ൗ 𝑛 𝜎 n = sample size Inferential Statistics Z − test for Comparing a Sample Mean xത vs Population Mean μ Example 7.2.1. A recent national survey found out that high school students spend an average of 6.8 hours per week watching television. A random sample of 36 high school students revealed that the mean number of hours they watched TV during the past week is 6.2 hours with a standard deviation of 0.5 hour. Assuming that the population is normally distributed test the hypothesis that the mean numbers of hours spend by the 36 high school students is not significantly lower than 6.8 hours. Use a 0.05 level of significance. Inferential Statistics Z − test for Comparing a Sample Mean xത vs Population Mean μ 1. Statement null and alternate hypotheses H0: The mean number of hours per week spent by the 36 high school students in watching television is not significantly lower than 6.8 hours. μ = 6.8 hours HA: The mean number of hours per week spent by the 36 high school students in watching television is significantly lower than 6.8 hours. μ < 6.8 hours Inferential Statistics Z − test for Comparing a Sample Mean xത vs Population Mean μ 2. Select a Level of significance α = 0.05 (one-tailed) Using the table for the Standard Normal  = 0.05 Cumulative Probability Failed to reject H0 Table, Critical Value Rejection region of H0 (Z) = -1.65 (since the alternative hypothesis made use of the inequality 30, σ is unknown but approximately equal to s). 𝑥ҧ − 𝜇 𝑍= ∙ 𝑛 𝜎 Where 𝑥ҧ = sample mean μ = population mean σ = population standard deviation n = sample size Inferential Statistics Z − test for Comparing a Sample Mean xത vs Population Mean μ 4. Formulate decision rule Decision Rule: Reject H0 if Z computed α, Failed to reject the null hypothesis. Table 7.2.1. will serve as your guide in testing hypothesis using p-value Method. Step 1. State the hypotheses and identify the claim. Step 2. Compute the test value. Step 3. Find the P-value. Step 4. Make the decision. Step 5. Summarize the results. Inferential Statistics p−Value Method for Hypothesis Testing Example 7.2.3. A researcher wishes to test the claim that the average cost of tuition and fees at a four-year public college is greater than $5700. She selects a random sample of 36 four-year public colleges and finds the mean to be $5950. The population standard deviation is $659. Is there evidence to support the claim at α = 0.05? Use the p-value method. Solution: Step 1. State the hypotheses and identify the claim. HO: μ = $5700 HA: μ > $5700 (claim) Inferential Statistics p−Value Method for Hypothesis Testing Step 2. Compute the test value. 𝑥ҧ − 𝜇 𝑧= ∙ 𝑛 𝜎 5,900 − 5,700 𝑧= ∙ 36 659 𝑧 = 2.28 Inferential Statistics p−Value Method for Hypothesis Testing Step 3. Find the P-value Using Standard Normal Cumulative Probability Table, find the corresponding area under the normal distribution for z = 2.28. It is 0.9887. Subtract this value for the area from 1.0000 to find the area in the right tail. 1.0000 − 0.9887 = 0.0113 Hence, the p-value is 0.0113. Inferential Statistics p−Value Method for Hypothesis Testing Step 4. Make the decision Since the p-value is less than 0.05, the decision is to reject the null hypothesis. See Figure 7.1.4. Figure 7.2.5. P-Value and α Value for Example 7.2.3. Source: Bluman, 2018 Inferential Statistics p−Value Method for Hypothesis Testing Step 5. Summarize the results There is enough evidence to support the claim that the tuition and fees at four-year public colleges are greater than $5700. Note: Had the researcher chosen α = 0.01, the null hypothesis would not have been rejected since the p-value (0.0113) is greater than 0.01. Inferential Statistics p−Value Method for Hypothesis Testing Activity 7.2.1. Answer the questions: (a) Is this a one- or two-tailed test? (b) What is the decision rule? (c) What is the value of the test statistic? (d) What is your decision regarding H0? (e) What is the p-value? Interpret it. 1. A sample of 36 observations is selected from a normal population. The sample mean is 49, and the population standard deviation is 5. Conduct the following test of hypothesis using the.05 significance level. 2. A sample of 36 observations is selected from a normal population. The sample mean is 12, and the population standard deviation is 3. Conduct the following test of hypothesis using the.02 significance level. 3. A sample of 36 observations is selected from a normal population. The sample mean is 21, and the population standard deviation is 5. Conduct the following test of hypothesis using the.05 significance levell Inferential Statistics p−Value Method for Hypothesis Testing Activity 7.2.1. For #4 and #5:(a) State the null hypothesis and the alternate hypothesis. (b) State the decision rule. (c) Compute the value of the test statistic. (d) What is your decision regarding H0? (e) What is the p-value? Interpret it. 4. The manufacturer of the X-15 steel-belted radial truck tire claims that the mean mileage the tire can be driven before the tread wears out is 60,000 miles. Assume the mileage wear follows the normal distribution and the standard deviation of the distribution is 5,000 miles. Crosset Truck Company bought 48 tires and found that the mean mileage for its trucks is 59,500 miles. Is Crosset’s experience different from that claimed by the manufacture at the.05 significance level? 5. The waiting time for customers at MacBurger Restaurants follows a normal distribution with a mean of 3 minutes and a standard deviation of 1 minute. At the Warren Road MacBurger, the quality-assurance department sampled 50 customers and found that the mean waiting time was 2.75 minutes. At the.05 significance level, can we conclude that the mean waiting time is less than 3 minutes? Inferential Statistics p−Value Method for Hypothesis Testing Performance Task 7. (Assignment No. 7.1.) 1. The XYZ Appliance Store issues its own credit card. The credit manager wants to find whether the mean monthly unpaid balance is more than P400. The level of significance is set 0.05. A random check of 172 unpaid balances revealed the sample mean P407 and the standard deviation of the sample is P38. Should the credit card manager conclude the population mean is greater than P400, or it is reasonable that the difference of P7 is due to chance? Inferential Statistics p−Value Method for Hypothesis Testing Performance Task 7. (Assignment No. 7.1.) 2. Heinz, a manufacturer of ketchup, uses a particular machine to dispense 16 ounces of its ketchup into containers. From many years of experience with the particular dispensing machine, Heinz knows the amount of product in each container follows a normal distribution with a mean of 16 ounces and a standard deviation of 0.15 ounce. A sample of 50 containers filled last hour revealed the mean amount per container was 16.017 ounces. Does this evidence suggest that the mean amount dispensed is different from 16 ounces? Use the.05 significance level. Inferential Statistics p−Value Method for Hypothesis Testing Performance Task 7. (Assignment No. 7.1.) a. State the null hypothesis and the alternate hypothesis. b. What is the probability of a Type I error? c. Give the formula for the test statistic d. State the decision rule. e. Determine the value of the test statistic. f. What is your decision regarding the null hypothesis? g. Interpret, in a single sentence, the result of the statistical test. LESSON 7.2. t Test for a Mean In most cases, however, the population standard deviation is unknown. Thus,  must be based on prior studies or estimated by the sample standard deviation, s. The population standard deviation in the following example is not known, so the sample standard deviation is used to estimate . Inferential Statistics 𝑡 Test for a mean ഥ to Population t Test for Comparing a Sample Mean 𝒙 Mean 𝝁 The t test is defined formally as follows. To find the value of the test statistic, we use the t distribution and revise formula as follows: with n – 1 is the degrees of freedom (df), Where: 𝑥ҧ − 𝜇 𝑥ҧ = the sample mean 𝑡= ∙ 𝑛 𝜇 = is the hypothesized population mean. 𝑠 s = is the sample standard deviation. n = is the number of observations in the sample. Inferential Statistics 𝑡 Test for a mean The t distribution is similar to the standard normal distribution in the following ways. 1. It is a continuous distribution 2. It is bell-shaped. 3. It is symmetric about the mean. 4. The mean, median, and mode are equal to 0 and are located at the center of the distribution. 5. The curve approaches but never touches the x axis. Inferential Statistics 𝑡 Test for a mean The t distribution differs from the standard normal distribution in the following ways. 1. The variance is greater than 1. 2. The t distribution is a family of curves based on the degrees of freedom, which is a number related to sample size. 3. As the number of degrees of freedom increases, the shape of the t distribution approaches that of the standard normal distribution. 4. The t distribution is flatter, or more spread out, than the standard normal distribution. 5. As the sample size increases, the t distribution approaches the normal distribution. Inferential Statistics 𝑡 Test for a mean Example 7.3.1. Find the critical t value for α = 0.05 with df = 16 for a right-tailed t test. Solution: Find the 0.05 column in the top row labeled One tail and 16 in the left-hand column. Where the row and column meet, the appropriate critical value is found; it is +1.746. See Figure 7.3.1. Figure 7.3.1. Finding the Critical Value for the t Test in t Distribution Table Inferential Statistics 𝑡 Test for a mean Example 7.3.2. Find the critical t value for α = 0.01 with df = 24 for a left-tailed t test. Solution: Find the critical value in the t Table on 0.01 column in the row labeled One-tail, and find 24 in the left column (df). The critical value is −2.492 since the test is left-tailed. Figure 7.3.2. Finding the Critical Value for the t Test in t Distribution Table Inferential Statistics 𝑡 Test for a mean Example 7.3.3. Find the critical values for α = 0.10 with df = 18 for a two-tailed t test. Solution: Find the 0.10 column in t Table on the row labeled Two-tails, and find 18 in the column labeled df. The critical values are +1.734 and −1.734. Figure 7.3.3. Finding the Critical Value for the t Test in t Distribution Table Inferential Statistics 𝑡 Test for a mean Assumptions for the t Test for a Mean When 𝛔 Is Unknown 1. The sample is a random sample. 2. Either n ≥ 30 or the population is normally distributed when n < 30. Recall: Five-Step Procedure for Testing a Hypothesis. Step 1. State null and alternate hypotheses Step 2. Select a level of significance Step 3. Identify the test statistic Step 4. Formulate a decision rule Step 5. Take a sample, arrive at decision Inferential Statistics t Test for Comparing a Sample Mean xത vs Population Mean μ Example 7.3.4. A leading brand of powdered orange juice claims that the Vitamin C content of their product is 60 mg per serving on the average. Assuming that the distribution is normally distributed. Test the claim, 9 samples yielded the following results. Determination No. 1 2 3 4 5 6 7 8 9 Vitamin C content(mg) 60.2 59.6 59.8 60.0 60.5 61 60.4 59.0 59.7 At α =.01, is there a significant difference in the mean Vitamin C content based on the manufacturers claim against the results of the analysis, done by the students? Inferential Statistics t Test for Comparing a Sample Mean xത vs Population Mean μ Solution: Step 1. State null and alternate hypotheses H0: The mean Vitamin C content of the 9 samples is not significantly different from 60 mg. 𝜇 = 60 𝑚𝑔 HA: The mean Vitamin C content of the 9 samples is significantly different from 60 mg. 𝜇 ≠ 60 𝑚𝑔 Inferential Statistics t Test for Comparing a Sample Mean xത vs Population Mean μ Solution: Step 2. Select a level of significance 𝛼 = 0.01 𝑡𝑤𝑜 − 𝑡𝑎𝑖𝑙𝑒𝑑 0.01 𝛼= = 0.005 2 𝑑𝑓 = 𝑛 − 1 𝑑𝑓 = 9 − 1 𝑑𝑓 = 8 𝐶𝑉: 𝑡0.005,8 = ±3.355 Inferential Statistics t Test for Comparing a Sample Mean xത vs Population Mean μ Step 3. Identify the test statistic Normally distributed, n < 30, population standard deviation is unknown Then, the formula is: 𝑥ҧ − 𝜇 𝑡= ∙ 𝑛 𝑠 Inferential Statistics t Test for Comparing a Sample Mean xത vs Population Mean μ Step 4. Formulate a decision rule Decision Rule: Reject H0 if t computed is < -3.355 or > 3.355, otherwise do not reject H0. Inferential Statistics t Test for Comparing a Sample Mean xത vs Population Mean μ Step 5. Take a sample, arrive at decision 𝑥ҧ − 𝜇 𝑡= ∙ 𝑛 𝑠 60.02 − 60.00 𝑡= ∙ 9 0.58 𝒕 = 𝟎. 𝟏𝟎 Since 0.10 < 3.355 do not reject Ho. There is not enough evidence to reject the claim of the manufacturer differs with the result of the students’ analysis Inferential Statistics ഥ𝟏 𝒗𝒔 ഥ t Test for Comparing Two Sample Means 𝒙 𝒙𝟐 1. The populations must be at least approximately normally distributed. 2. The two samples must be unrelated, that is, independent. 3. The population variances must be equal. Tests for equality of variances are taken up in advance courses in statistics. 4. The standard deviations for both populations must be known. 5. Populations have equal variances, pooled variance (sp2) will be computed. Inferential Statistics 𝒕 test for Comparing Two Sample Means 𝑥ҧ1 vs 𝑥ҧ2 ഥ𝟏 𝒗𝒔 ഥ t Test for Comparing Two Sample Means 𝒙 𝒙𝟐 The formula will be: 𝑥ҧ1 −𝑥ҧ2 𝑡= 1 1 𝑠𝑝 𝑛 +𝑛 1 2 𝑛1 − 1 𝜎1 2 + 𝑛2 − 1 𝜎2 2 𝑠𝑝 = 𝑛1 + 𝑛2 − 2 Inferential Statistics 𝒕 test for Comparing Two Sample Means 𝑥ҧ1 vs 𝑥ҧ2 Example 7.3.5. To find out whether a new drug will reduce the spread of cancer, 9 mice which have all reached an advance state of the disease are selected. Five mice receive the treatment and four do not. The survival periods, in months from the time that the experiment commenced are as follows: Treatment 2.4 5.3 1.4 4.6 0.9 No Treatment 1.9 0.8 2.8 3.7 At a 0.01 level of significance, is there evidence to say that the new drug is effective? Assume the two distributions to be normal with equal variances. Inferential Statistics t Test for Comparing a Sample Mean xത vs Population Mean μ Solution: Step 1. State null and alternate hypotheses H0: The mean survival time (in months) of mice treated with the new drug is not significantly longer than the mean survival time of mice not treated with the new drug. 𝜇1 = 𝜇2 HA: The mean survival time (in months) of mice treated with the new drug is significantly longer than the mean survival time of mice not treated with the new drug. 𝜇1 ≠ 𝜇2 Inferential Statistics t Test for Comparing a Sample Mean xത vs Population Mean μ Solution: Step 2. Select a level of significance 𝛼 = 0.01 𝑜𝑛𝑒 − 𝑡𝑎𝑖𝑙𝑒𝑑 𝑑𝑓 = (n1 + n2) – 2 𝑑𝑓 = (5 + 4) – 2 𝑑𝑓 = 7 𝐶𝑉: 𝑡0.01,7 = 2.998 Inferential Statistics t Test for Comparing a Sample Mean xത vs Population Mean μ Step 3. Identify the test statistic Normally distributed, n < 30, population standard deviation is unknown 𝑥ҧ1 − 𝑥ҧ2 𝑡= 1 1 𝑠𝑝 + 𝑛1 𝑛2 𝑛1 − 1 𝜎1 2 + 𝑛2 − 1 𝜎2 2 𝑠𝑝 = 𝑛1 + 𝑛2 − 2 Inferential Statistics t Test for Comparing a Sample Mean xത vs Population Mean μ Step 4. Formulate a decision rule Decision Rule : Reject H0 if computed t is > 2.998, otherwise failed to reject H0. Inferential Statistics t Test for Comparing a Sample Mean xത vs Population Mean μ Step 5. Take a sample, arrive at decision 𝑥ҧ1 = 2.92 𝑠1 2 = 3.79 𝑛1 2 = 5 𝑥ҧ2 = 2.30 𝑠2 2 = 1.54 𝑛2 2 = 4 2.92 − 2.30 5 − 1 (3.79) + 4 − 1 (1.54) 𝑡= 𝑠𝑝 = 1 1 5+4−2 1.68 + 5 4 𝑠𝑝 = 1.68 𝑡 = 0.55 Inferential Statistics t Test for Comparing a Sample Mean xത vs Population Mean μ Activity 7.3.1. 1. The following data on annual rates of return were collected from five stocks listed on the New York Stock Exchange (“the big board”) and five stocks listed on NASDAQ. Assume the population standard deviations are the same. At the.10 significance level, can we conclude that the annual rates of return are higher on the big board? NYSE NASDAQ 17.16 15.80 17.08 16.28 15.51 16.21 8.43 17.97 25.15 7.77 Inferential Statistics t Test for Comparing a Sample Mean xത vs Population Mean μ Activity 7.3.1. 2. Commercial Bank and Trust Company is studying the use of its automatic teller machines (ATMs). Of particular interest is whether young adults (under 25 years) use the machines more than senior citizens. To investigate further, samples of customers under 25 years of age and customers over 60 years of age were selected. The number of ATM transactions last month was determined for each selected individual, and the results are shown below. At the.01 significance level, can bank management conclude that younger customers use the ATMs more? Under 25 10 10 11 15 7 11 10 9 Over 60 4 8 7 7 4 5 1 7 4 10 5 Inferential Statistics t Test for Comparing a Sample Mean xത vs Population Mean μ Performance Task 7. (Assignment No. 7.2.) Advertisements by Sylph Fitness Center claim that completing its course will result in losing weight. A random sample of eight recent participants showed the following weights before and after completing the course. At the.01 significance level, can we conclude the students lost weight? Name Before After Hunter 155 154 Cashman 228 208 Mervine 151 147 Massa 162 157 Creola 211 196 Peterson 164 150 Redding 184 170 Poust 172 165 Inferential Statistics t Test for Comparing a Sample Mean xത vs Population Mean μ Performance Task 7. (Assignment No. 7.2.) (a) State the null hypothesis and the alternate hypothesis. (b) What is the critical value of t? (c) What is the computed value of t? (d) Interpret the result. What is the p-value? (e) What assumption needs to be made about the distribution of the differences? TAKE QUIZ Problem Solving 45 pts. Module References: Chi Square Distribution Table for degrees of dreedom 1-100. EasyCalculation. Retrieved from https://www.easycalculation.com/statistics/chisquare-table.php t Table. Retrieved from https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf Bluman, A. (2018). Elementary Statistics: A step by step approach. (10th Ed.). McGrawHill- Education. www.mhhe.com Glen. S, (2020). Hypothesis Testing. Retrieved From StatisticsHowTo.com: Elementary Statistics for the rest of us! https://www.statisticshowto.com/probability-and- statistics/hypothesis-testing/ Lind, D., Marchal, W., & Wathen, S. A. (2018). Statistical techniques in business & economics. (17th Ed). Mc.Graw-Hill Irwin. Do you have any questions? [email protected] THANK YOU! CREDITS: This presentation template Slidesgo was created by Slidesgo, including icons Flaticon by Flaticon, and infographics & images Freepik by Freepik. Please keep this slide for attribution.

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