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 = 4.26Adhesive Bonding strength (kg)typeAdhesive 1 10.2 11.8Adhesive 2 12.814.7Adhesive 3 7.29.8Table 2 Bonding strength data9.613.38.712.415.49.2

The question is about ANOVA Given : No. of adhesive types ( p ) = 3 No. of rep. in each sample ( r )…

Answered: Three adhesives are being analysed for…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Suppose a hypertension trial is mounted and 18 participants are randomly assigned to one of three comparison treatments. Each participant takes the assigned medication and his or her systolic blood pressure is recorded after 6 months on the assigned treatment. The data are shown in Table 7–58. Is there a difference in mean systolic blood pressure among treatments? Run the appropriate test at α = 0.05. TABLE 7–58 Systolic Blood Pressure by Treatment Standard Treatment Placebo New Treatment 124 134 114 111 143 117 133 148 121 125 142 124 128 150 122 115 160 128
Concrete blocks to be used in a certain application are supposed to have a mean compressive strength of 1500 MPa. Samples of size 1 are used for quality control. The compressive strengths of the last 40 samples are given in the following table. Sample Strength 1487 1463 3 1499 1502 1473 1520 1520 1495 1503 10 1499 11 1497 12 1516 13 1489 14 1545 15 1498 16 1503 17 1522 18 1502 1499 1484 19 20 21 1507 22 1474 23 1515 1533 1487 1518 24 25 26 27 1526 28 1469 29 1472 30 1512 1483 1505 31 32 33 1507 34 1505 1517 1504 35 36 37 1515 38 1467 39 1491 40 1488 Previous results suggest that a value of o = 15 is reasonable for this process. a. Using the value 1500 for the target mean u,and the values k = 0.5 and h = 4, construct a CUSUM chart. b. Is the process mean in control? If not, when is it first detected to be out of control?
A randomized experiment measured weight gain (in grams) of makes rats under six diets varyingby source of protein (beef, cereal, pork) and level of protein (low, high). Ten rats were assignedto each diet. Outline the design.
A producer of chocolate candy bars has developed an improved recipe for one of their most popular brands. They are conducting taste tests using randomly selected mall shoppers. One hundred shoppers were given the original recipe bar to taste and to rate it on a scale of 1 to 10. Another hundred shoppers were given the new recipe candy bar to taste and to rate it on a scale of 1 to 10. The results are in the table below for statistical analysis of variability between the two samples. All of these analyses are run wtih a 95% level of confidence. RATING, 1 to 10 Original Recipe New Recipe Sample Mean 7.8 8.1 Sample Standard Deviation 1.2 1.6 Use this information to answer the following questions. (a) Find the difference between these sample means. (b) Find the variance of the original recipe sample. (c) Find the variance of the new recipe sample. (d) Which test should be run to determine if there is evidence of a difference in consumer preferences between the two recipes?…
A survey is conducted to determine if there is a difference in the proportion of students, parents, and The conditions for inference are met. The chi-square test statistic is 0.19 and df = 2. What is the appropriate decision using a = 0.05? teachers who volunteer at least once a month. To investigate, a random sample of 45 students, 25 parents, and 12 teachers was selected from a large high school. The data are displayed in the table. Find the chi-square table here. Because the P-value is less than a = 0.05, the null hypothesis should be rejected. Student Parent Teacher Because the P-value is less than a = 0.05, the null hypothesis should not be rejected. Yes 24 12 Volunteer? No 21 13 Because the P-value is greater than a = 0.05, the null hypothesis should be rejected. The researcher would like to test these hypotheses: Because the P-value is greater than a = 0.05, the null hypothesis should not be rejected. Họ: There is no difference in the distribution of responses among these three…
The data in table is from a randomized block design represent blood pressure measurements from an experiment involving 4 age groups, each with 3 persons. The 3 persons within each age group were randomly assigned to drugs, A, B and C with one person per drug. This was done to keep the drug assignments balanced within age groups. Comparing the three drugs in manner that provides balance for or control for possible age effects. Consider the data given in table below and compute; a. Mean Square Due to Treatments (MSTR) b. Mean Square Due to Blocks (MSBL) Mean Square Due to Error (MSE) C. d. F-test Treatments Age Groups (Blocks) A B 1 70 72 2 76 84 3 82 86 4 90 92 Total 318 334 C Total Blocks 80 222 82 242 84 252 88 270 334 986
The authors of the paper "Statistical Methods for Assessing Agreement Between Two Methods of Clinical Measurement"† compared two different instruments for measuring a person's ability to breathe out air. (This measurement is helpful in diagnosing various lung disorders.) The two instruments considered were a Wright peak flow meter and a mini-Wright peak flow meter. Seventeen people participated in the study, and for each person air flow was measured once using the Wright meter and once using the mini-Wright meter. Subject Mini-WrightMeter WrightMeter Subject Mini-WrightMeter WrightMeter 1 512 494 10 445 433 2 430 395 11 432 417 3 520 516 12 626 656 4 428 434 13 260 267 5 500 476 14 477 478 6 600 557 15 259 178 7 364 413 16 350 423 8 380 442 17 451 427 9 658 650 (a) Suppose that the Wright meter is considered to provide a better measure of air flow, but the mini-Wright meter is easier to transport and to use. If the two types of meters produce different…
Don't need help with first part of a, got that figured out. Recall that small effects may be statistically significant if the samples are large. A study of small-business failures looked at 147 food-and-drink businesses. Of these, 103 were headed by men and 44 were headed by women. During a three-year period, 15 of the men's businesses and 6 of the women's businesses failed. (a) Find the proportions of failures for businesses headed by men (sample 1) and businesses headed by women (sample 2). These sample proportions are quite close to each other. p̂men = p̂women = Give the P-value for the z test of the hypothesis that the same proportion of women's and men's businesses fail. (Use the two-sided alternative.) The test is very far from being significant. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value = (b) Now suppose that the same sample proportions came from a sample of 30 times as large. That is, 180 out of…
The leader of two postpartum women’s support groups is interested in the depression levels of the women in her groups. She administers the Center for Epidemiologic Studies Depression Scale (CES-D) screening test to the members of her groups. The CES-D is a 20-question self-test that measures depressive feelings and behaviors during the previous week. The mean depression level from the screening test for the 10 women in the first group is μ₁ = 16; the mean depression level for the 14 women in the second group is μ₂ = 10. Without calculating the weighted mean for the combined group, you know that the weighted mean is:
A physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 54 men and 70 women to participate in the study. Each subject was required to step up and down a 6-inch platform. The pulse of each subject was then recorded. The following results were obtained. Two sample T for Men vs Women N Mean StDev SE Mean Men Women 98% CI for mu Men - mu Women (- 12.20, - 1.00) T-Test mu Men = mu Women (vs H2 O C. Ho: H1 = H2; Ha: H1 #H2 (b) Identify the P-value and state the researcher's conclusion if the level of significance was a = 0.01. What is the P-value? P-value =
A random sample of senior citizens living in a retirement village reported that they had an average of 1.42 face-to-face interactions per day with their neighbors. A random sample of those living in age-integrated communities reported 1.58 interactions per day. Is the difference between the two senior citizen groups significant? Please provide t critical, t obtained and your decision. Retirement Community 1 Retirement Community 2 Mean1 = 1.42 Mean2 = 1.58 s1 = 0.10 s2 = 0.78 N1 = 43 N2 = 37
Polyunsaturated fatty acids in the diet favorably affect several risk factors for cardiovascular disease. The principal dietary polyunsaturated fat is linoleic acid. Suppose we randomly selected twelve adults for our study. We took their blood pressure measurements (systolic blood pressure) at the start and recorded them for each adult. Then we placed required these adults to consume 23g/day of safflower oil which is high in linoleic acid for four weeks. At the end of the four weeks, we took another systolic blood pressure(SBP) measurement. The paired data is provided below: Subject 1 2 3 4 5 6 7 8 9 10 11 12 Baseline SBP 119.67 100.00 123.56 109.89 96.22 133.33 115.78 126.39 122.78 117.44 111.33 92.39 1-month SBP 117.33 98.78 123.83 107.67 95.67 128.89 113.22 121.56 126.33 110.39 107.00 93.06 Use the appropriate nonparametric test at alpha level of 0.05, to test whether the diet heavy in linoleic acid increases systolic blood pressure (SBP).

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