Below is a table of assembly times (in hours) for

Below is a table of assembly times (in hours) for 35 wind turbine blades produced in WindPower’s Secaucus, New Jersey assembly plant during the past two weeks.,a. Show the times in a grouped data frequency table, using the intervals 12 hours to under 14 hours, 14 hours to under 16 hours, 16 hours to under 18 hours, and so on.,b. Draw the histogram for the table that you pro-duced in part a.,c. Use the grouped data table to approximate the mean, the variance, and the standard deviation of the assembly times, and compare your results to the actual mean, variance and standard deviation of the raw data. (The mean of the raw data is 16.67; the variance is 7.5; the standard deviation is 2.74.)

Below is a sample of your bowling scores. (A perfect

Below is a sample of your bowling scores. (A perfect score is 300):,130, 99, 190, 40, 290, 130, 115,a. Determine the mean, the median and the mode for the data. Interpret each of these three measures of central tendency.,b. Summarize in a few words what these measures reveal about your bowling.

Below is a relative frequency table showing American household incomes

Below is a relative frequency table showing American household incomes as reported in the 2000 Census?,*To simplify your calculations, assume the last income class is $200,000 to $300,000. Notice that the classes become wider at higher levels of income. ,a. Draw the corresponding histogram. ,b. Estimate the mean of household incomes.

(a) n = 50, ˆp = 0.095 (b) n = 100,

(a) n = 50, ˆp = 0.095,(b) n = 100, ˆp= 0.095,(c) n = 500, ˆp = 0.095,(d) n = 1000, ˆp = 0.095,(e) Comment on the effect of sample size on the observed P-value of the test.

(a) Determine the critical value for a right-tailed test of

(a) Determine the critical value for a right-tailed test of a population mean at the ? = 0.01 level of signi?cance with 15 degrees of freedom.,(b) Determine the critical value for a left-tailed test of a population mean at the ? = 0.05 level of signi?cance based on a sample size of n = 20.,(c) Determine the critical values for a two-tailed test of a population mean at the ? = 0.05 level of signi?cance based on a sample size of n = 13.

1. To help the manufacturer get a clear picture of

1. To help the manufacturer get a clear picture of type I and type II error probabilities, draw a ? versus ? chart for sample sizes of 30, 40, 60, and 80. If ? is to be at most 1% with ? = 5%, which sample size among these four values is suitable?,2. Calculate the exact sample size required for ? = 5% and ? = 1%. Construct a sensitivity analysis table for the required sample size for ? ranging from 2,788 to 2,794 psi and ? ranging from 1% to 5%.,3. For the current practice of n = 40 and ? = 5% plot the power curve of the test. Can this chart be used to convince the manufacturer about the high probability of passing batches that have a strength of less than 2,800 psi?,4. To present the manufacturer with a comparison of a sample size of 80 versus 40, plot the OC curve for those two sample sizes. Keep an ? of 5%.,5. The manufacturer is hesitant to increase the sample size beyond 40 due to the concomitant increase in testing costs and, more important, due to the increased time required for the tests. The production process needs to wait until the tests are completed, and that means loss of production time. A suggestion is made by the production manager to increase ? to 10% as a means of reducing ?. Give an account of the benefits and the drawbacks of that move. Provide supporting numerical results wherever possible.,When a tire is constructed of more than one ply, the interply shear strength is an important property to check. The specification for a particular type of tire calls for a strength of 2,800 pounds per square inch (psi). The tire manufacturer tests the tires using the null hypothesis where ? is the mean strength of a large batch of tires. ,H0: ? ? 2,800 psi,From past experience, it is known that the population standard deviation is 20 psi.,Testing the shear strength requires a costly destructive test and therefore the sample size needs to be kept at a minimum. A type I error will result in the rejection of a large number of good tires and is therefore costly. A type II error of passing a faulty batch of tires can result in fatal accidents on the roads, and therefore is extremely costly. (For purposes of this case, the probability of type II error, ?, is always calculated at ? = 2,790 psi.) It is believed that ? should be at most 1%. Currently, the company conducts the test with a sample size of 40 and an ? of 5%.