Question 2The deflection temperature under load for two different types of plastic pipe is beinginvestigated. Two random samples of 12 pipe specimens are tested, and the deflectiontemperatures observed are as follows (in °F):Type-1 206 188 205 187 195 192 185 213 192 194 178 205Type-2 177 197 206 201 180 176 185 200 197 193 198 188Do the data support the claim that the deflection temperature under load for type 1 pipeexceeds that of type 2 at 5 % level of significance?

Claim: The defection temperature under load for type 1 pipe exceeds that of type 2 at 5% level of…

Answered: The deflection temperature under load…

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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Fiber Density. In the article “Comparison of Fiber Counting by TV Screen and Eyepieces of Phase Contrast Microscopy” (American Industrial Hygiene Association Journal, Vol. 63, pp. 756–761), I. Moa et al. reported on determining fiber density by two different methods. The fiber density of 10 samples with varying fiber density was obtained by using both an eyepiece method and a TV-screen method. A hypothesis test is to be performed to decide whether, on average, the eyepiece method gives a greater fiber density reading than the TV-screen method. a. identify the variable. b. identify the two populations. c. identify the pairs. d. identify the paired-difference variable. e. determine the null and alternative hypotheses. f. classify the hypothesis test as two tailed, left tailed, or right tailed.
Question 12 An analysis of variance experiment produced portions of a company ANOVA table Source of variations S.S Between groups Within groups d.f MS 25.08 Fral 3 a C. Total 117.72 79 N.B F(3,76)5% 2.724 and F3 79, 5% = 2.70 1. Specify the competing hypothesis in order to determine the population means. 2. Find the missing statistics a, b and c in the ANOVA table. 3. At the 5% significant level, what is the conclusion to the test? • Paragraph B IU A A I E E I I * x, E E E- & V G D I T 12pi
QUESTION 20 An environmental researcher wants to know whether the mean amount of sulfur dioxide in the air in UAE cities is less than 1.22 parts per billion (ppb). She tested the hypotheses Ho: 21.22 versus Ha: < 1.22 using a significance level a = 0.05. The p-value of the test is 0.045. If the true value of µ is 1.23, the conclusion results in Correct decision. Type I error. Both Type I and Type II errors. Type II error.
QUESTION 4 PART A- What is the standard error of the correlation coefficient for the association of extroversion and neuroticism in personalities? QUESTION 4 PART B- Carry out a formal t-test for the association between extroversion and neuroticism in personalities.xls. H0: The population correlation coefficient is 0. HA: The population correlation coefficient is not 0. What is the value of the t-statistic corresponding to this test? (Report the absolute value. The t-distribution is symmetric, thus it is easier to compare the absolute value of the t-statistic with the positive values in statistical table C). DATA IS ATTACHED.
Question 5 of 19 Compute the MSE for the following data. Period 1 2 3 4 5 6 13 Select the correct response. 12 11 Actual 492 470 485 493 498 492 14 Forecast (Error) 495 465 487 490 499 498 -5 -3 1 6 |Error] 3 LO 5 2 3 1 6 Error² 9 25 4 9 1 36 Error/Actual*100 0.60 1.06 0.41 0.60 0.20 1.21
Question 6 In a random sample of 12 adults each of them ate 100 g of dark chocolate and in another random sample of 15 adults each of them ate 100 g of mik cho plasma antioxidant capacity was measured and the summary statistics are given below. Sample mean Sample standard deviation Dark chocolate (DC) Milk chocolate (MC) 171.5 6.8 165.0 7.2 At 5% significance level do the data provide sufficient evidence that dark chocolate produces different mean level of total blood plasmà antioxidant c Assume that the population distributions are normal with equal variance. a. The alternative hypothesis is H1: Pparkc (/ +) HMIKC b. The test statistic t= (round at two decimal places) c. The degrees of freedom (d.f.) = d. The critical value of the test = (do not round up). than the critical value, we (accept/reject) e. As the test statistic is (less/greater) produce different mean level of total blood plasma antioxidant capa that dark chocolate (does/does not) -tailed test. f. The above test is a…
Question 4 of 7 Big babies: The National Health Statistics Reports described a study in which a sample of 315 one-year-old baby boys were weighed. Their mean weight was 25.6 pounds with standard deviation 5.3 pounds. A pediatrician claims that the mean weight of one-year-old boys is greater than 25 pounds. Do the data provide convincing evidence that the pediatrician's claim is true? Use the o 0.05 level of significance and the critical value method with the @ Critical Values for the Student's t Distribution Table. (a) State the appropriate null and alternate hypotheses. (b) Compute the value of the test statistic. (c) State a conclusion.
Question #3 Samples of body temperatures were collected for a group of women and a group of men. The summary statistics are given below: Women:n1 = 15 sample mean = 98.38 ̊F s1 = 0.45 ̊F Men:n2= 91 sample mean= 98.17 ̊F s2 = 0.65 ̊F 1. Use a 0.01 significance level to test the claim that women and men have different mean body temperatures using the p value method. Show all the steps of your hypothesis test including the p value: initial conclusion: a final conclusion about the original claim:
Question 9 Two independent samples are selected from two populations with means µg and u2, respectively. Suppose a 99% confidence interval for u1 - H2 is (-1.65, 21.65), a 95% confidence interval is (0.2, 19.8), and a 90% confidence interval is (1.775, 18.225). In testing Ho : H1 - H2 = 0 versus H1: H1 - H2 0, the p-value is between 0.20 and 1.00 between 0.10 and 0.15 between 0.05 and 0.10 between 0.01 and 0.05. between 0.15 and 0.20
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Question 20. part b. Nationally, patients who go to the emergency room wait an average of 7 hours to be admitted into the hospital. Do patients at rural hospitals have a lower waiting time? The 14 randomly selected patients who went to the emergency room at rural hospitals waited an average of 5.8 hours to be admitted into the hospital. The standard deviation for these 14 patients was 1.7 hours. What can be concluded at the the αα = 0.05 level of significance level of significance? C.)The test statistic ?, z or t = ____ (please show your answer to 3 decimal places.) D.)The p-value = ______ (Please show your answer to 4 decimal places.) E.) The p-value is ?, ≤ , > α

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