2.) The results of tension adhesion tests of 20 alloy samples are given below. Find the 95% two-sided confidence intervals on the mean tension adhesion. 19.8 11.9 17.6 18.5 15.4 11.4 19.5 10.1 15.4 13.6 14.1 11.9 15.8 7.5 8.8 12.7 14.9 11.4 16.7 7.9
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- Suppose that the LFPR in 1968 was 0.58 (or 58 percent). On the basis of the regression results given above, what is the mean LFPR in 1972? Establish a 95% confidence interval for the mean prediction.8.3 The table below shows a different random sampling of 20 cell phone models. Use this data to calculate a 93% confidence interval for the true mean SAR for cell phones certified for use in the US. As previously, assume that the population standard deviation is o = 0.337. Phone Model SAR Phone Model SAR Blackberry Pearl 8120 HTC Evo Design 4G 1.48 Nokia E71x 1.53 0.8 Nokia N75 0.68 HTC Freestyle LG Ally 1.15 Nokia N79 1.4 Sagem Puma Samsung Fascinate Samsung Infuse 4G Samsung Nexus S Samsung Replenish Sony W518a Walkman ZTE C79 1.36 1.24 LG Fathom 0.77 0.57 LG Optimus Vu 0.462 0.2 Motorola Cliq XT 1.36 0.51 Motorola Droid Pro 1.39 0.3 Motorola Droid Razr M 1.3 0.73 Nokia 7705 Twist 0.7 0.869 Solution A Solution B InterpretationThe cooling system in a nuclear submarine consists of an assembly of welded pipes through which a coolant is circulated. Specifications require that weld strength must exceed 150 psi. A random sample of 20 welds resulted in a sample mean of 153.7 psi and a sample standard deviation of 11.3 psi. What is the appropriate alternative hypothesis for the problem? O mean 150 O mean = 150 mean # 153.5
- If the observed concentration readings are 2.56, 2.67, 2.06, 2.68, and 2.43, determine 95% confidence interval for the mean concentration for the following two conditions: a) assuming variance (σ2) = 0.04 b) the variance is not knownIn order to estimate the difference between the average miles per gallon of two different models of automobiles, samples are taken and the following information is collected. Sample Size Sample Mean Sample Variance Model A a. 65 26 16 Model B 45 28 a 9 At 95% confidence, develop an interval estimate for the difference between the average miles per gallon for the two models. Is there conclusive evidence to indicate that one model gets a higher miles per gallon than the b. other? If yes, explain how you know this and which model is higher. If no, explain how you know that there is not enough evidence.200 people were randomly sampled and asked what they regularly eat for breakfast or lunch. Each person was identified as either a consumer or a non consumer of high-fiber cereals, and the number of calories consumed at lunch was measured and recorded. These data are summarized below; Consumer of high fiber cereals Non consumer of high fiber cereals η1 =41 η2 = 159 Mean 1 =603 Mean 2 =639 Stanadard deviation 1 = 110 Standard deviation 2 = 141 If the scientist claims that people who eat high fiber cereals for breakfast do consume on average fewer calories for lunch than people who don’t eat high fiber cereals for breakfast, and if it is true, high fiber cereal manufacturer will be able to claim another advantage of eating their products-potential weight reduction for dieter. REQUIRED Are there sufficient evidence at 5% significance level to support the above claim?
- An article shows data to compare several methods for predicting the shear strength for steel plate girders. Data for two of these methods, the Karlsruhe and Lehigh procedures, when applied nine times to a girder, are shown in the following table. We wish to determine whether there is any difference between the two methods. Assume the same standard deviations. (Note that we will perform unpaired comparison. We do not perform a paired test.) Karlsruhe Method Lehigh Method 1.21304 1.160384 1.251478 1.268834 1.218271 1.214254 1.232497 1.244961 1.142573 1.091621 1.286563 1.2297 1.239222 1.219269 1.161394 1.162844 1.169785 1.118203 1.231058 1.199258 1.233329 1.269486 1.153381 1.106666 1.2. For a sample standard deviation, use Excel function STDEV.S(), not STDEV.P(). Fill in the blanks below: Karlsruhe Method Lehigh Method x1 = x2 = s1 = s2 = n1 = n2 = 1.3. Do the data suggest that the two…Assume that Christmas trees sold in a farm have heights that are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. Find the value of the quartile Q3 (75% of trees are shorter than this height). 65.3 inches 64.3 inches 66.1 inches O 67.8 inchesIn the experiment, you repeated measurements 8 times for a resistor. After data processing, you determined that the standard deviation of the measurement is 2.80. What is the 95% confidence range of the result? O ±2.8N +1.94N +5.45N O ±8N
- Choose the appropriate statistical test. When computing, be sure to round each answer as indicated. A dentist wonders if depression affects ratings of tooth pain. In the general population, using a scale of 1-10 with higher values indicating more pain, the average pain rating for patients with toothaches is 6.8. A sample of 30 patients that show high levels of depression have an average pain rating of 7.1 (variance 0.8). What should the dentist determine? 1. Calculate the estimated standard error. (round to 3 decimals). [st.error] 2. What is thet-obtained? (round to 3 decimals). 3. What is the t-cv? (exact value) 4. What is your conclusion? Only type "Reject" or Retain"The void volume within a textile fabric affects comfort, flammability, and insulation properties. Permeability of a fabric refers to the accessibility of void space to the flow of a gas or liquid. An article gave summary information on air permeability (cm3/cm2/sec) for a number of different fabric types. Consider the following data on two different types of plain-weave fabric: Fabric Type Sample Size Sample Mean Sample Standard Deviation Cotton 10 51.41 0.75 Triacetate 10 132.15 3.59 Assuming that the porosity distributions for both types of fabric are normal, let's calculate a confidence interval for the difference between true average porosity for the cotton fabric and that for the acetate fabric, using a 95% confidence level. Before the appropriate t critical value can be selected, df must be determined: df = 0.5625 10 + 12.8881 10 2 (0.5625/10)2 9 + (12.8881/10)2 9 = 1.8092 0.1849…andomly selected students participated in an experiment to test their ability Co determine when one minute (60 seconds) has passed. Forty students yielded a a sample mean of 56.3 seconds. Assume that o = 6.5 seconds. %3D dents. Is it likely that their mean have an estimate that is reasonably close to 58 seconds?