etth Fate for waters produced using two different etching methods For each method, the etch rate for 15 wafers is measured. You conduct a two-sample t test to determine whether there is a significant difference in the average etch rate for the two methods You usea significance level of 0.05 Baseid on these results which of the following statements true? 103 102 10.1 10 9.9 98 9.7 96 Method 1 Method 2 Method t Test Method 2-Method 1 Assuming unequal variances Etch Rate ... eee es
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- Researchers investigated whether a new process for producing yarn could reduce the mean amount of volatile organic compounds (VOCs) emitted by carpet. From random samples of carpets, the researchers found the mean reduction of VOCs emitted by carpets made with yarn produced by the new process compared with that of carpets made with yarn produced by the traditional process was 13 parts per million (ppm). All conditions for inference were met, and the p-value for the appropriate hypothesis test was 0.095. Which of the following statements is the best interpretation of the p-value? (A) The probability that the null hypothesis is true is 0.095. (B) The probability that the alternative hypothesis is true is 0.095. (C) The probability of observing a mean reduction of 13 ppm is 0.095.(D) If the null hypothesis is true, the probability of observing a mean reduction of at least 13 ppm is 0.095. (E) If the null hypothesis is true, the probability of observing a mean reduction of at most 13 ppm…The makers of the Oral-B 3D toothbrush claims that using their brush will "significantly combat tooth decay" when compared with regular toothbrushes. College students have an average of 0.70 (s = 0.40) cavities filled in a given year. You randomly ask 100 students to use the Oral-B 3D toothbrush for one year. They have an average of 0.631 cavities filled in that year. Does the Oral-B 3D toothbrush combat tooth decay at a significance level of .05? Complete a hypothesis test of this study that the using the Oral-B 3D toothbrush reduces (e.g., a "directional" hypothesis) tooth decay. Be sure to include all six steps of a hypothesis test. Please show using excel also. ThanksA national organization has been working with utilities throughout the nation to find sites for large wind machines that generate electricity. Wind speeds must average more than 13 miles per hour (mph) for a site to be acceptable. Recently, the organization conducted wind speed tests at a particular site. To determine whether the site meets the organization's requirements, consider the test, Ho: μ =13 Ha: μ >13 where μ is the true mean wind speed at the site and a=.05 Suppose the observed significance level ( p-value) of the test is calculated to be p=0.4413 Interpret this result. Since the p-value greatly exceeds α = .05, there is strong evidence to reject the null hypothesis. Since the p-value exceeds α = .05, there is insufficient evidence to reject the null hypothesis. We are 55.87% confident that μ = 13. The probability of rejecting the null hypothesis is 0.4413.
- A random sample of 170 donations at a certain blood bank reveals that 84 were Type A blood. Does this suggest that the actual percentage of Type A donations differs from 45%, of the population having Type A blood? A. State H0 and Ha B. Test the hypothesis using the P-Value approach at a significance level of 3%, and express your decision. C Based on the given data, determine the extreme value without rejecting the H0.We want to test whether the mean fluride levels in the drinking water is better in city A or City B. In order to test this, a sample of 24 different water samples was taken from city A and a sample was taken from 23 different water samples in city B. City A has a sample average of 1.7 micrograms of fluride per deciliter of water and city B has a sample average 2.0 micrograms of lead per deciliter of water. Further, city A has a known standard deviation of A=.96 and city B has a known standard deviation of B=.89. Find the p-value for the following hypothesis test: Null: uA-uB=0 vs. Alternative: uA-uB<0.A medical researcher conducts a hypothesis test to test the claim that U.S. adult males have gained weight over the past 15 years. Assume that all the conditions for proceeding with a one-sample test on proportions have been met. The calculated test statistic is approximately 1.71 with an associated p-value of approximately 0.0436. Choose the conclusion that provides the best interpretation for the p-value at a significance level of a=0.05
- Given the SPSS output: (a) Test the null hypothesis of no difference in the true mean amount of sodium for the four provinces. Use α = 0.05. What conclusion can be drawn using Critical Value Approach? What is the p-value? What conclusion can be drawn using p-value approach? (b) Use Tukey's HSD method, if appropriate, to determine which means differ. Notes: You need to do all four-steps of hypotheses to perform the hypothesis test using both critical and p-value approach. Critical value should be found using F-distribution table, whereas p-value has already been reported in SPSS output. There is no need to calculate F-test statistic and p-value as they are part of SPSS output. -(c) In case you fail to reject the null hypotheses (i.e. the average sodium intake of four provinces), there is NO need to carry out Tukey's HSD test to investigate further where the difference occurs. You can also justify the reason for not doing HSD test from Tukey's HSD output.9. The alkaline batteries of a toy car were designed to last 35 hours, on average. However lots of customers complained that the batteries were lasting less than 35 hours. You decided to randomly sample 50 of the batteries. The mean of the sample was 29.3 hours, and the sample standard deviation 2.8 hours. Is there sufficient evidence to support the claim that the mean battery life is less than 35? Perform a hypothesis test at the 1% significance level using the p-value approach Step 1 Step 2 Step 3 Step 4A company that manufactures steel wires guarantees that the mean breaking strength (in kN) of the wires is greater than 50. They measure the strengths for a sample of wires and test H0 : μ ≤ 50 versus H1 : μ > 50. a) If a Type I error is made, what conclusion will be drawn regarding the mean breaking strength? b) If a Type II error is made, what conclusion will be drawn regarding the mean breaking strength?
- Before a company decides to purchase a large quantity of a solvent, it would like to see substantive evidence that the mean value of an impurity-forming substance is less than 1.0 ppb. What is the hypothesis to be tested here? What are the type I and type II statistical errors that can be made in this regard?Nationally, patients who go to the emergency room wait an average of 4 hours to be admitted into the hospital. Do patients at rural hospitals have a different waiting time? The 12 randomly selected patients who went to the emergency room at rural hospitals waited an average of 3.6 hours to be admitted into the hospital. The standard deviation for these 12 patients was 2.2 hours. What can be concluded at the the α = 0.01 level of significance level of significance? For this study, we should use ________ (t-test for population mean or z-test for population proportion) The null and alternative hypotheses would be: H0: ________ H1: ________ The test statistic (z or t) = _____ (please show your answer to 3 decimal places.) The p-value = _______ (Please show your answer to 4 decimal places.) The p-value is _______ α Based on this, we should __________ the null hypothesis. Thus, the final conclusion is that ... The data suggest the population…A survey was conducted to investigate the relationship between gender (male and females) and sector of employment (private, government and academia). Using the information provided, does a relationship exist between gender and employment sector at the 5% significance level? If the Chi-square test statistics = 0.529, what conclusion can be made? A. Since Chi-square test statistics > Chi-square critical value, do not reject H0 B. Since Chi-square test statistics < Chi-square critical value, do not reject H0 C. Since Chi-square test statistics < Chi-square critical value, Reject H0 D. Since Chi-square test statistics > Chi-square critical value, Reject H0