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- Consider the following sample observations on stabilized viscosity of asphalt specimens. 2,762 2,907 3,012 2,813 2,876 Suppose that for a particular application, it is required that true average viscosity be 3,000. Does this requirement appear to have been satisfied? State the appropriate hypotheses. (Use ? = 0.05.) H0: ? > 3,000 Ha: ? < 3,000 H0: ? = 3,000 Ha: ? ≠ 3,000 H0: ? ≠ 3,000 Ha: ? = 3,000 H0: ? < 3,000 Ha: ? = 3,000 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) t = P-value = What can you conclude? Reject the null hypothesis. There is sufficient evidence to conclude that the true average is viscosity differs from 3000.Do not reject the null hypothesis. There is sufficient evidence to conclude that the true average is viscosity differs from 3000. Reject the null hypothesis. There is not sufficient evidence to conclude that…a. Find the t values that form the boundaries of the critical region for a two-tailed test with ? = .05 for a sample size of n = 12. b. Repeat the above for a one-tailed test, with ? = .05.4.105 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…
- 1.3.2 Consider the output given below that was obtained using the One Proportion applet. Use information from the output to find the standardized statistic for a sample propor- tion value of 0.45. Probability of success (n): 0.30 Sample size (n): 25 Number of samples: 1000 O Animate Draw Samples Total = 1000 180 Mean = 0.301 SD = 0.091 120 60 0.08 0.16 0.24 0.32 0.40 0.48 0.56 0.64 Proportion of successWhich of the following is the best decision and conclusion based on the result below? H.: p = 0.10 Ha:p + 0.10 Critical Value: +1.645 Computed Test Statistics: z = 5.61 O a. Since the computed test statistics is less than the critical value, do not reject Ho. Therefore, we conclude that at 0.10 level of significance, there is enough evidence that the population proportion is different from 10%. O b. Since the computed test statistics is less than the critical value, do not reject Ho. Therefore, we conclude that at 0.10evel of significance, there is enough evidence that the population proportion is different from 10%. O c. Since the computed test statistics is greater than the critical value, reject Ho. Therefore, we conclude that at 0.10 level of significance, there is enough evidence that the population proportion is different from 10%. O d. Since the computed test statistics is less than the critical value, reject Ho. Therefore, we conclude that at 0.10 level of significance, there…11. State the null hypothesis and the alternative hypothesis for this situation: “To test the quality of the work of a commercial laboratory, duplicate analyses of a purified benzoic acid (68.8% C, 4.953% H) sample were requested. The means of the reported results are 68.5%C and 4.882% H. Is there any indication of systematic error in either analysis?"
- Researchers conducted a study investigating the number of viralinfections people contracted as a function of their stress over a 6-month period. Use the provided steps to conduct a one-factor between-subjects ANOVA that test whether there are any differences in number of infections across the four conditions.a. State the hypotheses. b. Determine the critical value for an alpha level of 0.05.4. Which of the following is used in model fit comparisons? A. Alkaike Information Criterion test B. The R-squared adjusted C. The Bayesian Information Criterion D. All of the above 5. Dr. Musantu conducted an estimation between y (dependent variable) and x (independent variable) in Stata. He obtained 0.332 coefficient of x and standard error of 0.021. What is the value of the t-statistic? A. 14.809 01392 B. 15.809 C. 16.809 D. None of the above 6. Is the above regressor (in question 8) significant at 5% (which gives 1.96 critical values) level of significance? A. No B. Yes 7. In a log-log regression model, what does the slope coefficient measure? A. The elasticity or percentage change of y with respect to a percentage change in x. B. The change in y which the model predicts for a unit change in x. C. The change in x which the model predicts for a unit change in y. D. 100 multiplied by the ratio y/x. 0.02111.33 Osteoporosis. Osteoporosis is a condition in which the bones become brittle due to loss of minerals. To diagnose os- teoporosis, an elaborate apparatus measures bone mineral den sity (BMD). BMD is usually reported in standardized form The standardization is based on a population of healthy young adults. The World Health Organization (WHO) criterion for osteoporosis is a BMD 2.5 or more standard deviations below the mean for young adults. BMD measurements in a popula- tion of people who are similar in age and sex roughly follow a Normal distribution.? (a) What percent of healthy young adults have osteoporosis by the WHO criterion? (b) Women aged 70 to 79 are, of course, not young adults. The mean BMD in this age is about -2 on the standard scale for young adults. Suppose that the standard deviation is the same as for young adults. What percent of this older population has osteoporosis!
- The calculated value for your Chi Square crosstabulation comes in at 9.888. The test has 4 degrees of freedom. Given a .05 alpha level, would you reject the null hypothesis? What is the minimum calculated value necessary to reach the .01 alpha level with 3 degrees of freedom in a Chi-Square cross-tabulation?17.34 (EX) Oligofructose and calcium absorption 1/6: Nondigestible oligosaccharides are known to stimulate calcium absorption in rats. A double-blind, randomized experiment investigated whether the consumption of oligofructose similarly stimulates calcium absorption in healthy male adolescents 14 to 16 years old. The subjects took a pill for nine days and had their calcium absorption tested on the last day. The experiment was repeated three weeks later. Some subjects received the oligofructose pill in the first round and then a pill containing sucrose (which served as a control). The order was switched for the remaining subjects. Here are the fractional calcium absorption data (in percent of intake) for 11 subjects: Subject Control 1 2 3 4 5 6 7 8 9 10 11 78.4 76.6 57.4 51.5 49.0 46.6 44.2 42.9 37.2 34.1 24.6 Oligofructose 62.0 95.1 46.5 49.4 89.7 43.8 50.3 51.6 66.6 52.7 54.0 Explain clearly why the matched pairs t test is the proper choice for this experimental design. O Because this…10.20 In a random sample of recent high school graduates, two characteristics were recorded (the average score and the number of correct answers for the SAT test). This information was classified as shown in Table 10.8. Is there a dependence between the number of correct answers on the SAT test and the average rankings statistically discernible at a level α = 0.01?