A researcher conducts a study with a sample of 41 observations to test the effectiveness of a new medication. The goal is to determine if the new medication changes the average recovery time from a specific illness, which is known to be 20 days with standard treatments. The null and alternative hypotheses are set as follows: Ho: μ = 20 H₂ µ ‡ 20 For each of the following t-values, calculate the p-value, compare it to x = 0.05, and state the decision about the null hypothesis. a) t = 1.684 b) t = 2.021 c) t=2.423
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- Evaluate the following grades: (Case 1) a grade of 220 points on an exam with u = 200 ando = 21. (Case 2) a grade of 90 points on an exam with µ = 80 and o = 8. * The better grade cannot be determined. The first grade is better. The two grades are equal statistically. The second grade is better.Arsenic-based additives in chicken feed have been banned by the European Union. Ifa restaurant chain finds significant evidence that the mean arsenic level of theirchickens is above 80 ppb (parts per billion), the chain will stop using that supplier ofchicken meat. The hypotheses are: H 0 : µ = 80H 1 : µ > 80 where µ represents the mean arsenic level in all chicken meat from that supplier.Samples from two different suppliers are analyzed, and the resulting p-values aregiven: Sample from Supplier A: p-value is 0.0003Sample from Supplier B: p-value is 0.3500 a) Interpret each p-value in terms of the probability of the results happening byrandom chance. b) Which p-value shows stronger evidence for the alternative hypothesis? c) Which supplier, A or B, should the chain get chickens from in order to avoid toohigh a level of arsenic?A question of medical interest is whether jogging leads to a reduction in systolic blood pressure. To learn about this question, eight non-jogging volunteers have agreed to begin a 1-month jogging program. At the end of the month, their blood pressures were determined and compared with earlier values. The data are presented in the Table 1. a. The appropriate hypotheses are: H0 :μ1−μ2 =0 versus Ha :μ1−μ2 /= (does not equal) 0 H0 :μ1−μ2 =0 versus Ha :μ1−μ2 <0 H0 :μ1−μ2 =0 versus Ha :μ1−μ2 >0 b. The value of test statistic is 1. 0.577 2. −0.577 3. 0.237 4. -0.237 c. The p−value of test is 1. 0.816 2. 0.408 3. 0.582 4. 0.291 d. At the significance level calculated in part (c), we conclude that jogging i. Leads to reduction in systolic blood pressureii. Does not lead to reduction in systolic blood pressure
- A pilot study was performed to investigate the effect of temperature, x (in degree Fahrenheit) on the electrical power consumed, y (in Watt) by an automotive factory. Other factors were kept constant and the data were collected from the study. The summary of the data are given as follows: n = 8, 2x = 401, 2301, x? = 22495 2y? = 666509, Σ > xy = 118652. %3D Compute the equation of the least squares regression line of y on x. Interpret the equation obtained in part (i). Predict the power consumption for a temperature of 65°F. (i) (ii) (iii) (iv) Compute the coefficient of determination and explain.For healthy adults, the mean pH level is u = 7.4 , a new drug for arthritis has been developed. However, it is thought that this drug may change blood pH. A random sample of 33 patients with arthritis took the drug for three months. Blood tests showed that the sample mean from these 33 patients was a pH of 8.0 and it is known that o = 2.3 . (Note, there is no unit for this type of data). Use α 0.05 Jevel of significance to test the claim that the drug has increased the mean pH level of the blood. What are the hypotheses in this problem? А: Но и 3D 7.4 HA u>7.4 US B: Ho u = 7.4 US НА и 8.0 D: Ho и %3 8.0 НА H < 8.0 US O A B.Fifty male subjects drank a measured amount x (in ounces) of a medication and the concentration y (in percent) in their blood of the active ingredient was measured 30 minutes later. The sample data are summarized by the following information: n = 50 Ex = 112.5 Ex? = 356.25 %3D Ey = 4.83 Ey = 0.667 Exy = 15.255 0 < x < 4.5 Or= 0.875 Or= 0.709 Or= -0.846 Or=0.460 Or= 0.965
- Poisoning by the pesticide DDT causes tremors and convulsions. In a study of DDT poisoning, researchers fed several rats a measured amount of DDT. They then made measurements on the rats’ nervous systems that might explain how DDT poisoning causes tremors. One important variable was the “absolutely refractory period,” the time required for a nerve to recover after a stimulus. This period varies normally. Measurements on four rats gave the following measurements (in milliseconds): 6, 1.7, 1.8, 1.9. Estimate the mean absolutely refractory period for all rats of this strain when subjected to the same treatment.Riboflavin (Vitamin B2) is determined in a cereal sample by measuring its fluorescence intensity(형광세기) in 5% acetic acid solution. A calibration curve was prepared by measuring the fluorescence intensities of a series of standards of increasing concentrations. The following data were obtained. Riboflavin (μg/mL) 0.000 0.100 0.200 0.400 0.800 Unknown sample Fluorescence intensity 0.0 5.8 12.2 22.3 43.3 15.4 (a) Use the method of least squares to obtain the best straight line through these five points (n=5). (b) Make a graph showing the experimental data and the calculated straight line. (c) An unknown sample gave an observed fluorescence intensity of 15.4. Calculate the concentration of Riboflavin (Vitamin B2) in the unknown sample (μg/mL). (d) Calculate the coefficient of determination (R2).For a population with µµ = 48 and ơo = 8, find the X value that corresponds to each of the following z-scores: A. 0.50 В. 2.00
- In a population with o = 6, a score of X = 48 corresponds to z = 1.25. The mean for this population is u =Let ?1 denote true average tread life for a premium brand of P205/65R15 radial tire, and let ?2 denote the true average tread life for an economy brand of the same size. Test H0: ?1 − ?2 = 5000 versus Ha: ?1 − ?2 > 5000 at level 0.01, using the following data: m = 45, x = 42,100, s1 = 2400, n = 45, y = 36,200, and s2 = 1800.Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value = State the conclusion in the problem context. Fail to reject H0. The data does not suggest that the difference in average tread life exceeds 5000.Reject H0. The data suggests that the difference in average tread life exceeds 5000. Fail to reject H0. The data suggests that the difference in average tread life exceeds 5000.Reject H0. The data does not suggest that the difference in average tread life exceeds 5000.Let ?1 denote true average tread life for a premium brand of P205/65R15 radial tire, and let ?2 denote the true average tread life for an economy brand of the same size. Test H0: ?1 − ?2 = 5000 versus Ha: ?1 − ?2 > 5000 at level 0.01, using the following data: m = 45, x = 42,100, s1 = 2400, n = 45, y = 36,200, and s2 = 1800. Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value = State the conclusion in the problem context. Fail to reject H0. The data does not suggest that the difference in average tread life exceeds 5000.Reject H0. The data suggests that the difference in average tread life exceeds 5000. Fail to reject H0. The data suggests that the difference in average tread life exceeds 5000.Reject H0. The data does not suggest that the difference in average tread life exceeds 5000.