The molecular weight of a particular polymer should fall between 750 and 1,035. Thirty samples of this material were analyzed with the results X-bar = 902 and s = 90. Assume that molecular weight is normally distributed.
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- A drug manufacturer forms tablets by compressing a granular material that contains the active ingredient and various fillers. The hardness of a sample from each batch of tablets produced is measured to control the compression process. The target value for 11.5. The hardness data for a random sample of 20 tablets from one large batch are given. the hardness is Н 11.627 11.374 11.383 11.477 = 11.613 11.592 11.715 11.570 A hypothesis test of Ho: μ = 11.2 Ha: μ # 11.2 11.493 11.458 11.485 11.623 11.602 11.552 11.509 11.472 11.360 11.463 11.429 11.531 where μ = the true mean hardness of the tablets using a = 0.05 has a P-value of 0.4494. Because the P-value of 0.4494 > a = = 0.05, we fail to reject Ho. We do not have convincing evidence that the true mean hardness of these tablets is different from 11.5. A 95% confidence interval for the true mean hardness measurement for this type of pill is (11.472, 11.561). Which is the following statements is not true with regards to the 95% confidence…Given that s.e.(b) = 61 and the estimate is 32, what is the t-stat?A sample from a normal population with u=50 and o =6 has a mean of M =48.20.If the sample mean corresponds to a z=-1.50,then how many scores are in the sample?
- A normal population has mean =μ22 and standard deviation =σ16 . Find the values that separate the middle 60% of the population from the top and bottom 20% . The values that separate the middle 60% of the population above from the top and bottom 20% are and . Enter the answers in ascending order and round to one decimal place.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. 2.00The National Institute of Standards and Technology (NIST) offers a wide variety of “standard reference materials” (SRMs) with accurately specified analyte concentrations. SRM 1951c is intended in part for “evaluating the accuracy of clinical procedures for the determination of total cholesterol . . . in human serum.” The certified total cholesterol is 6.244 ± 0.072 mmol/L. For the sake of this question, let’s treat 6.244 mmol/L as its “true” value (μ). We will use this standard to test a method with an intrinsic variability (for an experienced analyst) of σ = 0.131 mmol/L. In three trials, you obtained the following cholesterol concentrations (all in mmol/L): 6.476, 6.590, 6.338. 1. Showing all work, calculate a 95% confidence interval for your measured cholesterol concentration and write it as an intermediate result (with guard digits). Does your measurement differ from μ at 95% confidence? 2. Repeat part 1 for 90% confidence and explain any change in your conclusion. 3. Another…
- The average age for licensed drivers in a county is µ = 41.6, σ = 12, and the distribution is approximately normal. A county police officer was interested in whether the average age of drivers receiving speeding tickets differed from the average age of the driving population. She obtained a sample of N = 16 drivers with speeding tickets. The average age for this sample was M = 34.4 and alpha = 0.05. calculate statistical power for this study.The Specific Absorption Rate (SAR) for a cell phone measures the amount of radio frequency (RF) energy absorbed by the user's body when using the handset. Every cell phone emits RF energy. Different phone models have different SAR measures. To receive certification from the Federal Communications Commission (FCC) for sale in the United States, the SAR level for a cell phone must be no more than 1.5 watts per kilogram. A sample of 47 models was tested and the average of their Specific Absorption Rates (SARs) was found to be 1.18 watts per kilogram. Assume that the population standard deviation is 0.25 watts per kilogram. Construct a 90% confidence interval for the mean of the SARs for cell phones that received certification from FCC. Confidence interval:( , ) Interpretation: We are % confident that the mean of the SARs for cell phones that received certification from FCC is between watts per kilogram and watts per kilogram.Suppose the lengths of human pregnancies are normally distributed with μ = 266 days and o 16 days. Complete parts (a) and (b) below. = (a) The figure to the right represents the normal curve with μ = 266 days and o= 16 days. The area to the right of X = 300 is 0.0168. Provide two interpretations of this area. Provide one interpretation of the area using the given values. Select the correct choice below and fill in the answer boxes to complete your choice. (Type integers or decimals.) O A. The proportion of human pregnancies that last less than O B. The proportion of human pregnancies that last more than days is days is A X 266 300
- Find the normal curve and show solutionsA random sample from a normal population has:n = 8, X-bar = 16, and s = 3. Find the 95% PI for asingle new value from the population.The control department of a light bulb manufacturer randomly picks 4400 light bulbs from the production lot every week. The records show that, when there is no malfunction, the defect rate in the manufacturing process (due to imperfections in the material used) is 1%. When 1.25%o or more of the light bulbs in the sample of 4400 are defective, the control unit calls repair technicians for service. Answer the following. (If necessary, consult a list of formulas.) (a) Find the mean of p, where p is the proportion of defective light bulbs in a sample of 4400 when there is no malfunction. (b) Find the standard deviation of p. (c) Compute an approximation for P(p N 0.0125), which is the probability that the service technicians will be called even though the system is functioning properly. Round your answer to four decimal places.