1 Ultra high-performance concrete (UHPC) is a relatively new construction material that ischaracterized by strong adhesive properties with other materials. The following data are theadhesion measurements (in mJ/m²) for UHPC specimens adhered to steel: 107.1, 109.5,107.4, 106.8, 108.1. Assume that these sample observations were selected from a normaldistribution.a) Calculate a two-sided 95% confidence interval for the true average work of adhesion forUHPC adhered to steel using R.b) Predict the resulting work of adhesion value resulting from a single future replication ofthe experiment by calculating a 95% prediction interval and compare the width of thisinterval to the width of the CI from Part a) using R.

The data values are:107.1, 109.5, 107.4, 106.8, 108.1.Assume the sample observations are selected…

Answered: Ultra high-performance concrete (UHPC)…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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A group of students measure the length and width of a random sample of beans. They are interested in investigating the relationship between the length and width. Their summary statistics are displayed in the table below. All units, if applicable, are millimeters. Mean width: 7.555 Stdev width: 0.914 Mean height: 12.686 Stdev height: 1.634 Correlation coefficient: 0.8203 d) If the students are interested in using the height of the beans to predict the width, calculate the slope of this new regression equation. e) Write the equation of the best-fit line that can be used to predict bean widths. Use x to represent height and y to represent width.
Shear strength measurements derived from unconfined compression tests for two types of sols gave the results shown in the following table (measurements in tons per square foot). Do the soils appear to differ with respect to the variation of the shear strength, at the 10% significance level? Soil Soil Type Турe l II sample size 10 sample 1.57 1.92 mean sample std dev 0.17 0.22 Let the strength measurement of soil type I be the first population and the strength measurement of soil type II be the second. a. Choose the answer : 1) Alternative Hypothesis: The population variation of the shear strength for soil type l is > that of soil type I. 2) Alternative Hypothesis: The population variation of the shear strength for soil type I is < that of soil type II. 3) Alternative Hypothesis: The sample variation of the shear strength for soil type l is not equal to that of soil type II. 4) Alternative Hypothesis: The population variation of the shear strength for soil type l is not equal to that of…
Elasticity is an important parameter of kernmantle ropes (used in climbing, caving, mountaineering, etc.). One of the ways of describing it is static elongation (measured under a resting load of 80 kg). Static elongation was measured for two random rope samples of two different types of weaving. type-1 6.34 6.16 6.32 6.21 6.36 6.43 6.02 6.21 6.47 6.21 6.49 6.21 6.16 6.19 6.29 6.33 6.51 type-2 6.38 6.51 6.49 6.44 6.46 6.43 6.52 6.32 6.33 6.32 6.31 6.34 6.13 6.42 (b) Find the P-value for this test. O 0.01 < P - value < 0.025 0.1 < P - value < 0.25 0.05 < P – value < 0.1 0.0025 < P – value < 0.005 (c) Construct a 95% confidence interval for the difference in the mean static elongation of ropes. Round your answers to three decimal places (e.g. 98.765).
21 smooth/yellow 42 smooth/light-yellow 19 smooth/white 22 wrinkled/yellow 39 wrinkled/light-yellow 20 wrinkled/white Calculate the Chi Sqaure statistic of the data shown above:
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?
The strength of concrete depends, to some extent, on the method used for drying. Two different drying methods showed the following results for independently tested specimens (measurements in psi): Do the methods appear to produce concrete with different mean strengths? Use � = 0.05 .
For screened coke, the porosity factor is measured by the difference in weight between dry and soaked coke. A certain supply of coke is claimed to have a porosity factor of 1.5 kilograms. Ten samples are tested, resulting in amean porosity factor of 1.9 kilograms and a variance of 0.04. Is there sufficient evidence to indicate that the coke is more porous than is claimed? Use and assume the porosity measurements are approximately normally distributed.
4.105 INDEX FINGER MUSCLE FATIGUE AND PISTOL FIRING. Refer to the Human Factors (November 2019) study of the effects of finger strength and muscle fatigue on the ability to shoot a pistol, Exercise 2.80 (p. 86). Recall that muscle strength was measured (in Newtons) by the maximum voluntary isometric contraction (MVIC) of the index finger. The researchers found that when at rest, expert male shooters have a mean MVIC of 160.3 N and a standard deviation of 19.6 N. However, after repeated shooting sessions, the expert shooters have a mean MVIC of 133.5 N and a standard deviation of 19.6 N. Assume that the distribution of MVIC strength scores in both scenarios is approximately normal. a. When at rest, what is the probability that an expert shooter has an MVIC value of 120 N or less? b. After repeated shooting sessions, what is the probability that an expert shooter has an MVIC value of 120 N or less? c. What MVIC value will be exceeded by 75% of the expert shooters when at rest? d. Repeat…
A company produces chips to sell for commercial electronic products. A specific product type is required to have a maximum variance of 0.001 sq. cm. The finished products are to undergo quality testing before shipping. A sample of sixteen chips are measured using a precise instrument having the following observations: 1.63 1.65 1.71 1.64 1.69 1.62 1.63 1.70 1.64 1.59 1.66 1.63 1.65 1.57 1.66 1.69 Assuming that the data is normally distributed, use a level of significance of 0.025 to test whether the lot should be shipped and that it conforms to the set specifications of the company a.What are the hypothesis statements? b.Which hypothesis test is appropriate for the study? c.What is the critical value? type in up to 2 decimal places d.What is the conclusion of the study?
A manufacturing company produces bearings. One line of bearings is specified to be 1.64 centimeters (cm) in diameter. A major customer requires that the variance of the bearings be no more than .001 cm². The producer is required to test the bearings before they are shipped, and so the diameters of 16 bearings are measured with a precise instrument, resulting in the following values. Assume bearing diameters are normally distributed. Use the data and a = 0.01 to test to determine whether the population of these bearings is to be rejected because of too high a variance. 1.68 1.62 1.63 1.70 1.66 1.63 1.65 1.71 1.64 1.69 1.57 1.64 1.59 1.66 1.63 1.65 Appendix A Statistical Tables (Round your answer to 2 decimal places, e.g. 15.25.) The value of the test statistic is and we reject the null hypothesis ⇒
An article includes the accompanying data on compression strength (lb) for a sample of 12-oz aluminum cans filled with strawberry drink and another sample filled with cola. Beverage Sample Size Sample Mean Sample SD Strawberry Drink 10 535 24 Cola 10 559 15 Does the data suggest that the extra carbonation of cola results in a higher average compression strength? Base your answer on a P-value. (Use ? = 0.05.) State the relevant hypotheses. (Use ?1 for the strawberry drink and ?2 for the cola.) H0: ?1 − ?2 = 0Ha: ?1 − ?2 ≥ 0H0: ?1 − ?2 = 0Ha: ?1 − ?2 ≠ 0 H0: ?1 − ?2 = 0Ha: ?1 − ?2 < 0H0: ?1 − ?2 = 0Ha: ?1 − ?2 > 0 Calculate the test statistic and determine the P-value. (Round your test statistic to one decimal place and your P-value to three decimal places.) t = P-value = State the conclusion in the problem context. Reject H0. The data suggests that cola has a higher average compression strength than the strawberry drink.Reject H0. The…
The lowest level of significance to reject the null hypothesis of no linear association between blood pressure and age is: OA: 0.003 OB: 0.05 OC: 0.0002 OD: 0.0001 OE: 0.04

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