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Hypothesis testing Hypothesis testing A study showed that 30% of men are coffee drinkers compared to 20% for women! Possible Interpretation !????? 1. True difference that also exists in the population from which we drew our sample 2. Difference might be due to chance 3. Difference is due to bias Choosing a significance test Bias can be ruled out if we feel confident that our study design and research implementation techniques were sound The distinction between differences due to chance or real variation need significance testing A significance test Estimates the likelihood that an observed study result is due to chance It is used to find out whether a study result which is observed in a sample can be considered as a result which exists in the population from which the sample was drawn If you are measuring an association between two variables, you determine how likely it is that your results could have occurred by chance (i.e: occurred due to sampling variation) A significance test If it is unlikely (< 5%) that your result occurred by chance, you reject the chance explanation and accept that there is a real difference The difference is statistically significant If it is likely (> 5%) that your result occurred by chance, you cannot conclude that a real difference exists. Therefore the difference is not statistically significant Null hypothesis The assumption that in the total population no real difference exists between groups (no real association exists between variables) Example: Males do not smoke more than females Statistical difference A statistically significant difference does not necessarily mean that the relationship is an important one Developing Null and Alternative Hypotheses 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 H0 , is a tentative assumption about a population parameter. The alternative hypothesis, denoted by Ha, is the opposite of what is stated in the null hypothesis. A Summary of Forms for Null and Alternative Hypotheses about a Population Mean The equality part of the hypotheses always appears in the null hypothesis. In general, a hypothesis test about the value of a population mean must take one of the following three forms (where 0 is the hypothesized value of the population mean). H0: > 0 Ha: < 0 H0: < 0 Ha: > 0 H0: = 0 Ha: 0 Example: Metro EMS Null and Alternative Hypotheses A major west coast city provides one of the most comprehensive emergency medical services in the world. Operating in a multiple hospital system with approximately 20 mobile medical units, the service goal is to respond to medical emergencies with a mean time of 12 minutes or less. The director of medical services wants to formulate a hypothesis test that could use a sample of emergency response times to determine whether or not the service goal of 12 minutes or less is being achieved. Example: Metro EMS Null and Alternative Hypotheses Hypotheses Conclusion and Action H0: The emergency service is meeting the response goal; no follow-up action is necessary. Ha: The emergency service is not meeting the response goal; appropriate follow-up action is necessary. Where: = mean response time for the population of medical emergency requests. Type I and Type II Errors Since hypothesis tests are based on sample data, we must allow for the possibility of errors. A Type I error is rejecting H0 when it is true. A Type II error is accepting H0 when it is false. The person conducting the hypothesis test specifies the maximum allowable probability of making a Type I error, denoted by and called the level of significance. Generally, we cannot control for the probability of making a Type II error, denoted by . Statistician avoids the risk of making a Type II error by using “do not reject H0” and not “accept H0”. Example: Metro EMS Type I and Type II Errors Conclusion Population Condition True status H0 True H0 False ( ) ( ) Accept H0 (Conclude Correct Conclusion Reject H0 (Conclude Type I rror Type II Error Correct Conclusion P-Value It is the likelihood or probability of observing a result by chance A p<0.05 means that you would observe a difference in your data only 5 times or less in every 100 samples examined Choosing a significance test Before applying any significance test, state the null hypothesis in relation to the data to which the test is being applied. This will enable you to interpret the results of the test The Use of p-Values The p-value is the probability of obtaining a sample result that is at least as unlikely as what is observed. The p-value can be used to make the decision in a hypothesis test by noting that: • if the p-value is less than the level of significance , the value of the test statistic is in the rejection region. • if the p-value is greater than or equal to , the value of the test statistic is not in the rejection region. Reject H0 if the p-value < . Statistical significance and causation An association that is statistically significance does not necessarily imply the existence of a causal relation. It calls for further investigation to find out whether a causal relationship does exist The Steps of Hypothesis Testing Determine the appropriate hypotheses. Select the test statistic for deciding whether or not to reject the null hypothesis. Specify the level of significance for the test. Use to develop the rule for rejecting H0. Collect the sample data and compute the value of the test statistic. a) Compare the test statistic to the critical value(s) in the rejection rule, or b) Compute the p-value based on the test statistic and compare it to to determine whether or not to reject H0 . Two-Tailed Tests about a Population Mean: Large-Sample Case (n > 30) Hypotheses H0: = Ha: Test Statistic Known Unknown z= x 0 / n Rejection Rule t = x s/ n 0 Reject H0 if |z | > |z1- /2 | Reject H0 if |t | > |t1- /2 | Example: Glow Toothpaste Two-Tailed Tests about a Population Mean: Large n The production line for Glow toothpaste is designed to fill tubes of toothpaste with a mean weight of 6 ounces. Periodically, a sample of 30 tubes will be selected in order to check the filling process. Quality assurance procedures call for the continuation of the filling process if the sample results are consistent with the assumption that the mean filling weight for the population of toothpaste tubes is 6 ounces; otherwise the filling process will be stopped and adjusted. Example: Glow Toothpaste Two-Tailed Tests about a Population Mean: Large n A hypothesis test about the population mean can be used to help determine when the filling process should continue operating and when it should be stopped and corrected. Hypotheses H0: = Ha: Rejection Rule assuming a .05 level of significance, Two-Tailed Test about a Population Mean: Large n Assume that a sample of 30 toothpaste tubes provides a sample mean of 6.1 ounces and standard deviation of 0.2 ounces. Let n = 30, H0: = Ha: ≠ = 6.1, x s = .2 ounces x 0 6.1 6 t= = = 2.74 s / n .2 / 30 Since 2.74 > Critical value, we reject H0. The p-value .0062 is less than = .05, so H0 is rejected Conclusion: We are 95% confident that the mean filling weight of the toothpaste tubes is not 6 ounces. The filling process should be stopped and the filling mechanism adjusted . Confidence Interval Approach to a Two-Tailed Test about a Population Mean Select a simple random sample from the population and use the value of the sample mean x to develop the confidence interval for the population mean . If the confidence interval contains the hypothesized value 0, do not reject H0. Otherwise, reject H0. Example: Glow Toothpaste Confidence Interval Approach to a Two-Tailed Hypothesis Test The 95% confidence interval for is s x t1 / 2 n For example, if the confidence interval is (6.01 to 6.2) Since the hypothesized value for the population mean, 0 = 6, is not in this interval, the hypothesis-testing conclusion is that the null hypothesis, H0: = 6, can be rejected. Tests about a Population Mean: Small-Sample Case (n < 30) Test Statistic t= Unknown x 0 s/ n This test statistic has a t distribution with n - 1 degrees of freedom. Rejection Rule One-Tailed H0: H0: H0: = Two-Tailed Reject H0 if | t| > t Reject H0 if |t| > t p -Values and the t Distribution 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. However, we can still use the t distribution table to identify a range for the p-value. An advantage of computer software packages is that the computer output will provide the p-value for the t distribution. One-Tailed Tests about a Population Mean: Large-Sample Case (n > 30) Hypotheses H0: Ha: H0: Ha: Test Statistic Known z= or x 0 / n Unknown t = x s/ n 0 Rejection Rule Reject H0 if |z| > |z1-| Accept H0 if |z| < |z| One-Tailed Test about a Population Mean: Large n Let n = 40, x= 13.25 minutes, s = 3.2 minutes x 13.25 12 = = 2.47 s / sncan3.be 2 / used 40 to (The sample standard deviation t= estimate the population standard deviation .) Since 2.47 > critical value, we reject H0. Conclusion: We are 95% confident that Metro EMS is not meeting the response goal of 12 minutes; appropriate action should be taken to improve service. Example: Highway Patrol One-Tailed Test about a Population Mean: Small n A State Highway Patrol periodically samples vehicle speeds at various locations on a particular roadway. The sample of vehicle speeds is used to test the hypothesis H0: < 65. The locations where H0 is rejected are deemed the best locations for radar traps. At Location F, a sample of 16 vehicles shows a mean speed of 68.2 mph with a standard deviation of 3.8 mph. Use an = .05 to test the hypothesis. Example: Highway Patrol One-Tailed Test about a Population Mean: Small n Let n = 16, = 68.2 mph, s = 3.8 mph x = 16-1 = 15, t = 1.753 = .05, d.f. x 0 68.2 65 t= = = 3.37 s / n 3.8 / 16 Since 3.37 > 1.753, we reject H0. Conclusion: We are 95% confident that the mean speed of vehicles at Location F is greater than 65 mph. Location F is a good candidate for a radar trap. Summary of Test Statistics to be Used in a Hypothesis Test about a Population Mean Yes s known ? Yes n > 30 ? No Yes Use s to estimate s s known ? Yes x z= / n No x t= s/ n x z= / n No Popul. approx. normal ? No Use s to estimate s x t= s/ n Increase n to > 30 A Summary of Forms for Null and Alternative Hypotheses about a Population Proportion The equality part of the hypotheses always appears in the null hypothesis. In general, a hypothesis test about the value of a population proportion p must take one of the following three forms (where p0 is the hypothesized value of the population proportion). H0: p > p0 H0: p < p0 H0: p = p0 Ha: p < p0 Ha: p > p0 Ha: p p0 Tests about a Population Proportion: Large-Sample Case (np > 5 and n(1 - p) > 5) Test Statistic z= p = where: p p0 p p0 (1 p0 ) n Rejection Rule One-Tailed H0: p p H0: p p H0: p=p Two-Tailed Reject H0 if z > z Reject H0 if z < -z Reject H0 if |z| > z Example: NSC Two-Tailed Test about a Population Proportion: Large n For a Christmas and New Year’s week, the National Safety Council estimated that 500 people would be killed and 25,000 injured on the nation’s roads. The NSC claimed that 50% of the accidents would be caused by drunk driving. A sample of 120 accidents showed that 67 were caused by drunk driving. Use these data to test the NSC’s claim with = 0.05. Example: NSC Two-Tailed Test about a Population Proportion: Large n Hypothesis H0: p = .5 Ha: p .5 Test Statistic p0 (1 p0 ) .5(1 .5) p = = = .045644 n 120 z= p p0 p = (67 /120) .5 = 1.278 .045644 Example: NSC Two-Tailed Test about a Population Proportion: Large n Rejection Rule Reject H0 if z < -1.96 or z > 1.96 Conclusion Do not reject H0. For z = 1.278, the p-value is .201. If we reject H0, we exceed the maximum allowed risk of committing a Type I error (p-value > .050). Hypothesis Testing and Decision Making In many decision-making situations the decision maker may want, and in some cases may be forced, to take action with both the conclusion do not reject H0 and the conclusion reject H0. In such situations, it is recommended that the hypothesis-testing procedure be extended to include consideration of making a Type II error. Determining Differences Between Groups Introduction t-test is used for numerical data when comparing the means of two groups (mean of income, age, weight...etc for males and females) 2 is used for categorical data when comparing proportions of events occurring in two groups (prevalence rate in rural and urban) Student’s t-test or t-test Is the observed difference between means of two groups statistically ? significant Comparisons Involving Means Estimation of the Difference Between the Means of Two Populations: Independent Samples Hypothesis Tests about the Difference between the Means of Two Populations: Independent Samples Inferences about the Difference between the Means of Two Populations: Matched Samples Inferences about the Difference between the Proportions of Two Populations: Estimation of the Difference Between the Means of Two Populations: Independent Samples Point Estimator of the Difference between the Means of Two Populations Sampling Distribution x1 x2 Interval Estimate of Large-Sample Case Interval Estimate of Small-Sample Case Point Estimator of the Difference Between the Means of Two Populations Let 1 equal the mean of population 1 and 2 equal the mean of population 2. The difference between the two population means is 1 - 2. To estimate 1 - 2, we will select a simple random sample of size n1 from population 1 and a simple random sample of size n2 from population 2. Let equal the mean of sample 1 and equal the x2 meanx1 of sample 2. The point estimator of the difference between the means of the populations 1 and 2 is x1 x2 Sampling Distribution of x1 x2 Properties of the Sampling Distribution of x1 x2 Standard Deviation x1 x2 = 12 n1 22 n2 where: 1 = standard deviation of population 1 2 = standard deviation of population 2 n1 = sample size from population 1 n2 = sample size from population 2 Interval Estimate of 1 - 2: Large-Sample Case (n1 > 30 and n2 > 30) Interval Estimate with 1 and 2 Known where: x1 x2 z / 2 x1 x2 1 - is the confidence coefficient Interval Estimate with 1 and 2 Unknown x1 x2 z / 2 sx1 x2 where: sx1 x2 s12 s22 = n1 n2 Example: Par, Inc. Interval Estimate of 1 - 2: Large-Sample Case 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. Example: Par, Inc. Interval Estimate of 1 - 2: Large-Sample Case Sample Statistics Sample #1 Sample Size Mean Standard Dev. Sample #2 Par, Inc. Rap, Ltd. n1 = 120 balls n2 = 80 balls x1 = 235 yards x2= 218 yards s1 = 15 yards s2 = 20 yards Example: Par, Inc. Point Estimate of the Difference Between Two Population Means 1 = mean distance for the population of Par, Inc. golf balls 2 = mean distance for the population of Rap, Ltd. golf balls Point estimate of 1 - 2 = x1 x2 = 235 - 218 = 17 yards. Point Estimator of the Difference Between the Means of Two Populations Population 1 Par, Inc. Golf Balls Population 2 Rap, Ltd. Golf Balls 1 = mean driving 2 = mean driving distance of Par golf balls distance of Rap golf balls m1 – 2 = difference between the mean distances Simple random sample of n1 Par golf balls Simple random sample of n2 Rap golf balls x1 = sample mean distance for sample of Par golf ball x2 = sample mean distance for sample of Rap golf ball x1 - x2 = Point Estimate of m1 – 2 Example: Par, Inc. 95% Confidence Interval Estimate of the Difference Between Two Population Means: Large-Sample Case, 1 and 2 Unknown Substituting the sample standard deviations for the population standard deviation: x1 x2 z / 2 12 22 (15) 2 ( 20) 2 = 17 1. 96 n1 n2 120 80 = 17 + 5.14 or 11.86 yards to 22.14 yards. We are 95% confident that the difference between the mean driving distances of Par, Inc. balls and Rap, Ltd. balls lies in the interval of 11.86 to 22.14 yards. Interval Estimate of 1 - 2: Small-Sample Case (n1 < 30 and/or n2 < 30) Interval Estimate with 2 Known x1 x2 z / 2 x1 x2 where: x1 x2 1 1 = ( ) n1 n2 2 Interval Estimate of 1 - 2: Small-Sample Case (n1 < 30 and/or n2 < 30) Interval Estimate with 2 Unknown x1 x2 t / 2 sx1 x2 where: sx1 x2 1 1 = sp ( ) n1 n2 2 2 2 ( n 1 ) s ( n 1 ) s 1 2 2 sp 2 = 1 n1 n2 2 Example: Specific Motors Specific Motors of Detroit has developed a new automobile known as the M car. 12 M cars and 8 J cars (from Japan) were road tested to compare miles-per-gallon (mpg) performance. The sample statistics are: Sample Size Mean Standard Deviation Sample #1 M Cars n1 = 12 cars x1 = 29.8 mpg s1 = 2.56 mpg Sample #2 J Cars n2 = 8 cars x2= 27.3 mpg s2 = 1.81 mpg Example: Specific Motors Point Estimate of the Difference Between Two Population Means 1 = mean miles-per-gallon for the population of M cars 2 = mean miles-per-gallon for the population of J cars Point estimate of 1 - 2 = mpg. x1 x2= 29.8 - 27.3 = 2.5 Example: Specific Motors 95% Confidence Interval Estimate of the Difference Between Two Population Means: Small-Sample Case We will make the following assumptions: The miles per gallon rating must be normally distributed for both the M car and the J car. The variance in the miles per gallon rating must be the same for both the M car and the J car. Using the t distribution with n1 + n2 - 2 = 18 degrees of freedom, the appropriate t value is t.025 = 2.101. We will use a weighted average of the two sample variances as the pooled estimator of 2. Example: Specific Motors 95% Confidence Interval Estimate of the Difference Between Two Population Means: Small-Sample Case 2 2 2 2 ( n 1 ) s ( n 1 ) s 11 ( 2 . 56 ) 7 ( 1 . 81 ) 1 2 2 s2 = 1 = = 5. 28 n1 n2 2 12 8 2 x1 x2 t.025 1 1 1 1 s ( ) = 2. 5 2.101 5. 28( ) n1 n2 12 8 2 = 2.5 + 2.2 or .3 to 4.7 miles per gallon. We are 95% confident that the difference between the mean mpg ratings of the two car types is from .3 to 4.7 mpg (with the M car having the higher mpg). Hypothesis Tests About the Difference Between the Means of Two Populations: Independent Samples Hypotheses H0: 1 - 2 < 0 H0: 1 - 2 > 0 H0: 1 - 2 = 0 Ha: 1 - 2 > 0 Ha: 1 - 2 < 0 Ha: 1 - 2 0 Test Statistic Large-Sample z= ( x1 x2 ) ( 1 2 ) 12 n1 22 n2 Small-Sample t= ( x1 x2 ) ( 1 2 ) s2 (1 n1 1 n2 ) Example: Par, Inc. Hypothesis Tests About the Difference Between the Means of Two Populations: Large-Sample Case 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. Example: Par, Inc. Hypothesis Tests About the Difference Between the Means of Two Populations: Large-Sample Case Sample Statistics Sample #1 Sample Size Mean Standard Dev. Par, Inc. n1 = 120 balls x1 = 235 yards s1 = 15 yards Sample #2 Rap, Ltd. n2 = 80 balls x2 = 218 yards s2 = 20 yards Example: Par, Inc. Hypothesis Tests About the Difference Between the Means of Two Populations: Large-Sample Case Can we conclude, using a .01 level of significance, that the mean driving distance of Par, Inc. golf balls is greater than the mean driving distance of Rap, Ltd. golf balls? 1 = mean distance for the population of Par, Inc. golf balls 2 = mean distance for the population of Rap, Ltd. golf balls Hypotheses H0: 1 - 2 < 0 Ha: 1 - 2 > 0 Example: Par, Inc. Hypothesis Tests About the Difference Between the Means of Two Populations: Large-Sample Case Rejection Rule Reject H0 if z > 2.33 z= ( x1 x2 ) ( 1 2 ) s12 s22 n1 n2 (235 218) 0 17 = = = 6.49 2 2 2.62 (15) (20) 120 80 Conclusion Reject H0. We are at least 99% confident that the mean driving distance of Par, Inc. golf balls is greater than the mean driving distance of Rap, Ltd. golf balls. Example: Specific Motors Hypothesis Tests About the Difference Between the Means of Two Populations: Small-Sample Case Can we conclude, using a .05 level of significance, that the miles-per-gallon (mpg) performance of M cars is greater than the miles-pergallon performance of J cars? 1 = mean mpg for the population of M cars 2 = mean mpg for the population of J cars Hypotheses H0: 1 - 2 < 0 Ha: 1 - 2 > 0 Example: Specific Motors Hypothesis Tests About the Difference Between the Means of Two Populations: Small-Sample Case Rejection Rule Reject H0 if t > 1.734 (a = .05, d.f. = 18) Test Statistic where: t= ( x1 x2 ) ( 1 2 ) sp 2 (1 n1 1 n2 ) 2 2 ( n 1 ) s ( n 1 ) s 1 2 2 sp 2 = 1 n1 n2 2 Mean Heights of Males and Females in X School Gender Number of Mean Height students in cm Standard deviation Males 60 156 3.1 Females 52 154 2.8 Steps to follow Calculate the t-value Use the t-table and find out the tabulated t Interpret the results Mean Heights of Males and Females in Hala School Gender Number of students Mean Height in cm Standard deviation Males 60 156 3.1 Females 52 154 2.8 Using a T-Table 1- Decide the significance level (p value) you want to use (in health studies we usually 0.05) Do not forget that: if p = or < 0.05 then we reject the null hypothesis and accept the alternative one concluding the presence of statistically significant difference if p > 0.05 then we accept the null hypothesis and conclude that there is no statistically significant difference. Using a t-Table 2- Determine the degrees of freedom (sum of observations for the two groups minus 2) 3- Look for the t-value belonging to the chosen significance level 4- Compare tabulated t value and the calculated value 5-Apply the rule of accepting or rejecting the null hypothesis Using a t-Table 6 - if the calculated t-value is larger than the table value then p < than .05 and you end up rejecting the null hypothesis and concluding the presence of difference 7 - if the calculated t-value is smaller than the table value, then p > .05 and you end up accepting the null hypothesis and concluding the absence of difference Example: Express Deliveries Inference About the Difference Between the Means of Two Populations: Matched Samples di ( 7 6... 5) d = = = 2. 7 n 10 2 76.1 ( di d ) sd = = = 2. 9 n 1 9 d d 2. 7 0 t= = = 2. 94 sd n 2. 9 10 Conclusion Reject H0. There is a significant difference between the mean delivery times for the two services. Inferences About the Difference Between the Proportions of Two Populations p1 p2 Sampling Distribution of Interval Estimation of p1 - p2 Hypothesis Tests about p1 - p2 Sampling Distribution of p1 p2 Expected Value E ( p1 p2 ) = p1 p2 Standard Deviation p1 p2 = p1 (1 p1 ) p2 (1 p2 ) n1 n2 Distribution Form If the sample sizes are large (n1p1, n1(1 - p1), n2p2, and n2(1 - p2) are all greater than or equal to 5), the sampling distribution of p1 pcan be approximated by a 2 normal probability distribution. Interval Estimation of p1 - p2 Interval Estimate Point Estimator of p1 p2 s p1 p2 = p1 p2 z / 2 p1 p2 p1 (1 p1 ) p2 (1 p2 ) n1 n2 Example: MRA MRA (Market Research Associates) is conducting research to evaluate the effectiveness of a client’s new advertising campaign. Before the new campaign began, a telephone survey of 150 households in the test market area showed 60 households “aware” of the client’s product. The new campaign has been initiated with TV and newspaper advertisements running for three weeks. A survey conducted immediately after the new campaign showed 120 of 250 households “aware” of the client’s product. Does the data support the position that the advertising campaign has provided an increased awareness of the client’s product? Example: MRA Point Estimator of the Difference Between the Proportions of Two Populations p1 p2 = p1 p2 = 120 60 =. 48. 40 =. 08 250 150 p1 = proportion of the population of households “aware” of the product after the new campaign p2 = proportion of the population of households “aware” of the product before the new campaign p1= sample proportion of households “aware” of the product after the new campaign sample proportion of households “aware” of the p= 2 product before the new campaign Example: MRA Interval Estimate of p1 - p2: Large-Sample Case For = .05, z.025 = 1.96: . 48. 40 1. 96 . 48(.52) . 40(. 60) 250 150 .08 + 1.96(.0510) .08 + .10 or -.02 to +.18 Conclusion At a 95% confidence level, the interval estimate of the difference between the proportion of households aware of the client’s product before and after the new advertising campaign is -.02 to +.18. Hypothesis Tests about p1 - p2 Hypotheses H0: p1 - p2 < 0 Ha: p1 - p2 > 0 Test statistic z= ( p1 p2 ) ( p1 p2 ) p1 p2 s p1 p2 = p (1 p )(1 n1 1 n2 ) Point Estimator of p1 p2 where p1 = p2 where: n1 p1 n2 p2 p= n1 n2 Example: MRA Hypothesis Tests about p1 - p2 Can we conclude, using a .05 level of significance, that the proportion of households aware of the client’s product increased after the new advertising campaign? p1 = proportion of the population of households “aware” of the product after the new campaign p2 = proportion of the population of households “aware” of the product before the new campaign Hypotheses H0: p1 - p2 < 0 Ha: p1 - p2 > 0 Example: MRA Hypothesis Tests about p1 - p2 Rejection Rule Reject H0 if z > 1.645 Test Statistic 250(. 48) 150(. 40) 180 p= = =. 45 250 150 400 s p1 p2 = . 45(.55)( 1 1 ) =. 0514 250 150 (. 48. 40) 0 . 08 z= = = 1.56 . 0514 . 0514 Conclusion Do not reject H0.