You conduct a 1-way ANOVA with 7 groups and find a statistically significant result (using a = 0.02). You intend to follow this up with a post-hoc analysis. In particular, you plan to examine all pairwise contrasts. You retain the same nominal significance level for all tests. Round your answers to three decimal places, if necessary. How many pairwise contrasts will you need to examine? What is the probability of making at least one Type I error (Familywise Error Rate)?
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- Which of these scenarios means that a pretest-posttest design with a non-equivalent control group is not suitable? Select all that apply. O a. All of the available participants will be receiving the intervention at the same time. O b. Authors want to examine for any trends in the data over time. O c. The participants were placed into the experimental and control groups by natural circumstances rather than random assignment. O d. Authors want to compare averages between two groups before and after the intervention using an ANOVA.Suppose a study reported that the average persin watched 3.37 hours of television per day. a random sample of 15 people gave the number of hours of television watched per day shown below. at the 1% significance level, do the data provide sufficent evidence to conclude that the amount of television watched per day last year by the average person is greater than the value reported in the study? what is the test statistic value for this problem? a.0.28 b.-38 c.none of the above d. 0.58 e.-4.295 f..04 g. 1.94 . the proper conclusion for this problem is? a. reject the null hypothesis we have sufficent evidence to prove the average number of hours people watch tv is more than 3.37 hours. b. Do not reject the null hypothesis we have sufficient evidence to prove the average number of hours of TV watched is greater than 3.37 hours c. Reject the null hypothesis and claim the average number of hours people watch TV is less than 3.37 because the P value is near zero d. Do not reject the null…The Ahmadi Corporation wants to increase the productivity of its line workers. Four different programs have been suggested to help increase productivity. Twenty employees, making up a sample, have been randomly assigned to one of the four programs and their output for a day's work has been recorded. You are given the results below. At 95% confidence, conduct the ANOVA test to determine whether there is a significant difference in the mean output under the four programs. Program A Program B Program C Program D 150 150 185 175 130 120 120 135 180 160 145 110 Part of the ANOVA table is shown below. Source of Variation Sum of Squares 8,750 Between Groups Within Groups 16,350 Total 220 190 180 175 d.f. 19 150 120 130 175 Mean Square
- You are conducting a study to see if the accuracy rate for fingerprint identification is significantly different from 0.51. You use a significance level of a = 0.001. Ho:p = 0.51 H₁:p 0.51 You obtain a sample of size n = 373 in which there are 169 successes. What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = The p-value is... O less than (or equal to) a greater than a This p-value leads to a decision to... O reject the null O accept the null O fail to reject the null As such, the final conclusion is that... O There is sufficient evidence to warrant rejection of the claim that the accuracy rate for fingerprint identification is different from 0.51. O There is not sufficient evidence to warrant rejection of the claim that the accuracy rate for fingerprint identification is different from 0.51. O The sample data support the claim that the accuracy rate for fingerprint identification is different from 0.51. O There is not sufficient sample…A national organization that conducts research on the cost and quality of health care in the U.S. reported that, in 2012, U.S. families spent an average of $9,590 on health care expenses. Suppose you decide to test whetherthe average in 2015 is greater than the average in 2012. After conducting the appropriate statistical test, you find a P-value of 0.022. If the level of significance is 0.05, which of the following is the best interpretation of the P-value? a. The P-value of 0.022 indicates that there is a 2.2% chance that the 2015 average is greater than the average amount spent in 2012.b. The P-value of 0.022 provides weak evidence that the average in 2015 average is greater than the average amount spent in 2012.c. The P-value of 0.022 provides strong evidence that the 2015 average is greater than the average amount spent in 2012.d. The P-value of 0.022 indicates that there is a very low probability that the 2015 average is different than average amount spent in 2012.Acne is a common skin disease that affects most adolescents and can continue into adulthood. A study compared the effectiveness of three acne treatments and a placebo, all in gel form, applied twice daily for 12 weeks. The study's 517 teenage volunteers were randomly assigned to one of the four treatments. Success was assessed as clear or almost clear skin at the end of the 12 week period. The results of the study can be seen in the table below. Using the appropriate statistical test, determine if there is significant evidence that the four treatments perform differently. If so, how do they compare.
- Since an instant replay system for tennis was introduced at a major tournament, men challenged 1413 referee calls, with the result that 426 of the calls were overturned. Women challenged 742 referee calls, and 228 of the calls were overturned. Use a 0.05 significance level to test the claim that men and women have equal success in challenging calls. Identify the test statistic. z= ? (Round to two decimal places as needed.) Part 3 Identify the P-value. P-value= ? (Round to three decimal places as needed.) Part 4 What is the conclusion based on the hypothesis test? The P-value is ▼ greater than or less than the significance level of α=0.05, so ▼ fail to reject or reject the null hypothesis. There ▼ is not sufficient or is sufficient evidence to warrant rejection of the claim that women and men have equal success in challenging calls. Part 5 b. Test the claim by constructing an appropriate confidence interval. The 95% confidence…Choose the appropriate statistical test. When computing, be sure to round each answer as indicated. A dentist wonders if depression affects ratings of tooth pain. In the general population, using a scale of 1-10 with higher values indicating more pain, the average pain rating for patients with toothaches is 6.8. A sample of 30 patients that show high levels of depression have an average pain rating of 7.1 (variance 0.8). What should the dentist determine? 1. Calculate the estimated standard error. (round to 3 decimals). [st.error] 2. What is thet-obtained? (round to 3 decimals). 3. What is the t-cv? (exact value) 4. What is your conclusion? Only type "Reject" or Retain"In general, why does using a repeated-measures ANOVA increase statistical power compared to using a one-way between-groups ANOVA? Group of answer choices: a.MSbetween is lower b.MSerror is higher c.MSerror is lower d.MSbetween is higher
- A publisher reports that 69 % of their readers own a particular make of car. A marketing executive wants to test the claim that the percentage is actually under the reported percentage. A random sample of 200 found that 60% of the readers owned a particular make of car. Is there sufficient evidence at the 0.10 level to support the executive's claim? Step 5 of 7: Identify the value of the level of significance.A statistician changes her level of significance from .001 to .05. What impact will this change have on her risk of making a Type I and Type II error? Is the change in the level of significance a good decision? Why or why not?I In 2000, 6.1% of people used marijuana. This year, a company wishes to use their employment drug screening to test a claim. They take a simple random sample of 2848 job applicants and find that 145 individuals fail the drug test for marijuana. They want to test the claim that the proportion of the population failing the test is lower than 6.1%. Use .10 for the significance level. Round to three decimal places where appropriate. Hypotheses: Ho:p=6.1%Ho:p=6.1% H1:p<6.1%H1:p<6.1% Test Statistic: z = Critical Value: z = p-value: Conclusion About the Null: Reject the null hypothesis Fail to reject the null hypothesis Conclusion About the Claim: There is sufficient evidence to support the claim that the proportion of the population failing the test is lower than 6.1% There is NOT sufficient evidence to support the claim that the proportion of the population failing the test is lower than 6.1% There is sufficient evidence to warrant rejection of the claim that the…