Sample 1 2 3 4 5 6 7 8 9 10 Nonconformities 19 21 15 23 13 21 15 24 20 19 n this data: ) The total number of nonconformities found = 11 Sample 11 12 13 14 15 16 17 890 18 19 20 3) The average number of nonconformities per sample= C) The upper 3-sigma control limit for the c-chart = D) The lower 3-sigma control limit for the c-chart = E) Is the process in control? (Yes/No) Nonconformities he 37 16428 17 29 25 15 11 19
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- How many subjects participated in an independent-samples t-test if a researcher reports,t (20) = 3.68, alpha = .05, two-tail?Three adhesives are being analysed for their impact on the bonding strength of paper in a pulp and paper mill. The adhesives are each randomly applied to four batches. The data is shown in Table 2. Here, the three treatments are the adhesives (p=3) and the number of replications for each treatment is 4 (r=4). Is there a difference among the adhesives in terms of the mean bonding strength? Test at the 5% significance level. Given F0.05,2,9 = 4.26 Adhesive Bonding strength (kg) type Adhesive 1 10.2 11.8 Adhesive 2 12.8 14.7 Adhesive 3 7.2 9.8 Table 2 Bonding strength data 9.6 13.3 8.7 12.4 15.4 9.2We wish to test if there is a difference in the median weights for yellow and brown M&M's. A random sample of both types of M&M's is selected and weighed. The data was analyzed in Statdisk and with the following results. Total Num Values: 59 Rank Sum 1: Rank Sum 2: 829.5000 940.5000 780 Mean : St Dev: 65.49809 Test Statistic, z: 0.7557 Critical z: ±1.959962 Fail to Reject the Null Hypothesis Of these two choices (Wilcoxon signed-rank for matched pairs or Wilcoxon signed-rank for two independent samples) what type of test should have been conducted? What can be concluded? O The Wilcoxon signed-rank for matched pairs should have been conducted. There is insufficient evidence to conclude that there is a difference in the median weight of yellow and brown M&M's. The Wilcoxon signed-rank for matched pairs should have been conducted. There is sufficient evidence to conclude that there is a difference in the median weight of yellow and brown M&M's. The Wilcoxon rank-sum for two independent…
- Is narcissism a more common personality trait today than it was a few decades ago? It is known that the mean population score on the Narcissistic Personality Inventory (NPI) for students attending University of South Alabama around 20 years ago was μ= 15 (Twenge, 2010). Interested in the narcissism levels of students in the year 2020, a researcher administers the NPI to a random sample of 25 University of Alabama sophomores this Spring term. The mean NPI score from the researcher’s sample of sophomores is M = 16.5, with s = 3.4. 1. Write the null and alternative hypotheses in symbols. Possible symbols for your answer: H0, H1, μ, M, σ. 2. Calculate the standard error. 3. Find the critical value for the test statistic, assuming alpha = .05 (Use largest [i.e., most conservative] value if exact value not given in the chart) a) 2.064 b) 1.96 c) 1.98 d) 2.000Is narcissism a more common personality trait today than it was a few decades ago? It is known that the mean population score on the Narcissistic Personality Inventory (NPI) for students attending University of South Alabama around 20 years ago was μ= 15 (Twenge, 2010). Interested in the narcissism levels of students in the year 2020, a researcher administers the NPI to a random sample of 25 University of Alabama sophomores this Spring term. The mean NPI score from the researcher’s sample of sophomores is M = 16.5, with s = 3.4. 1. Find the obtained (i.e., computed) test statistic for a sample (n=25) with a mean of 16.5 2. Make a statistical decision about the null. Will you reject or fail to reject the null based on your sample data? 3. Justify your decision about the null.Five samples of a ferrous-type substance were used to determine if there is a difference between a laboratory chemical analysis and an X-ray fluorescence analysis of the iron content. Each sample was split into two subsamples and the two types of analysis were applied, with the accompanying results. Assuming that the populations are normal, test at the 0.02 level of significance whether the two methods of analysis give, on the average, the same result. Click here to view the sample analyses. Click here to view page 1 of the table of critical values of the t-distribution. Click here to view page 2 of the table of critical values of the t-distribution. Let sample 1 be the X-ray fluorescence results and let sample 2 be the laboratory chemical results. State the null and alternative hypotheses. Ho: HD H1: HD + Identify the critical region. Select the correct choice below and fill in the answer box(es) to complete your choice. (Round to three decimal places as needed.) O A. t> В. t O C. t<
- 6.59. Control charts for and s have been maintained on a process and have exhibited statistical control. The sample size is n = 6. The control chart parameters are as follows: I Chart s Chart UCL = 708.20 UCL = 3.420 Center line = 706.00 Center line = 1.738 LCL = 703.80 LCL = 0.052 (a) Estimate the mean and standard deviation of the process. (b) Estimate the natural tolerance limits for the process. (c) Assume that the process output is well modeled by a normal distribution. If specifications are 703 and 709, estimate the fraction nonconforming. (d) Suppose the process mean shifts to 702.00 while the standard deviation remains constant. What is the probability of an out-of-control signal occur- ring on the first sample following the shift? (e) For the shift in part (d), what is the probability of detecting the shift by at least the third subsequent sample?The yield of alfalfa from a random sample of six test plots is 1.4, 1.6, 0.9, 1.9, 2.2,and 1.2 tons per acre. Assume the data can be looked upon as a sample from anormal population. Test at the 0.05 level of significance whether this supports thecontention that the average yield for this kind of alfalfa is 1.5 tones per acre.15. Samples of pages were randomly selected from three different novels. The Flesch Reading Ease scores were obtained from each page, and the TI -83/84 Plus calculator results from analysis of variance are given below. Use a 0.01 significance level to test the claim that the three books have the same mean Flesch Reading Ease score. One-way ANOVA F = 2.63411669884 P = 0.0877903995 Factor Df = 2 SS = 389.080864 MS = 194.540432 Error Df = 31 SS = 2289.47844 MS = 73.8541431 Sxp = 8. 59384333 Identify the null hypothesis, alternate hypothesis, test statistic, P-value, conclusion about the null hypothesis, and conclusion that addresses the original claim.
- In an analysis of variance study (ANOVA) , individuals were separated to two groups to test the effect of a specific medicine on blood pressure. Each group consisted of 5 individuals, and blood pressure was recorded for each . Choose the correct statement : 15 Select one: 22 a. Factor : blood pressure, Treatments : group 1, group 2, Response variable : Medicine type O b. Factor: Medicine type, Treatments : group 1, group 2, response variable : blood pressure 29 c. Factor : group 1, group 2, Treatments : blood pressure, Response variable : Medicine type 36 O d. Factor : Medicine type, Treatments: blood pressure , Response variable : group 1,group 2 Finis ge Next page CTIVITY NEXT ACTIVITY Jump to... Midterm 1 étv Ps Lr Ai TA hypertensions trial is mounted and 12 participants are randomly assigned to receive either a new medication or a placebo. Each participant takes assigned medication and their systolic blood pressure (SBP) is recorded after 6 months on the assigned medication the data are shown in table 7-9 is there a difference in the mean SBP between treatment? Run the appropriate test at a =0.05Five samples of a ferrous-type substance were used to determine if there is a difference between a laboratory chemical analysis and an X-ray fluorescence analysis of the iron content. Each sample was split into two subsamples and the two types of analysis were applied, with the accompanying results. Assuming that the populations are normal, test at the 0.05 level of significance whether the two methods of analysis give, on the average, the same result. Determine the test statistic t=?