A local tennis pro-shop strings tennis rackets at the tension (pounds per square inch) requested by the customer. Recently a customer made a claim that the pro-shop consistently strings rackets at lower tensions, on average, than requested. To support this claim, the customer asked the pro shop to string 6 new rackets at 68 psi. Suppose the two-tailed P-value for the test described above (obtained from a computer printout) is 0.07. Find the conclusion for the test at the 5% significance level.

The approximate​ Z-score that corresponds to a right tail area of 0.42 is

An engineer claims the population mean of a certain batch process is 400 grams. To check his claim he takes a random sample of 36 batches and he is satisfied with his claim if his t-value lies in the range (t0.90, t0.10). Should he be satisfied with his claim if his random sample has a sample mean of 430 grams and a sample standard deviation of 90 grams?

Fill in the blank: A nonparametric test has less OA. power O B. type II error rates OC. standard deviations OD. means OE. None of the above than a t test.

Based on the image attached, it shows monthly wind speed data (measured in m/s) was obtained. The researcher claimed that the average wind speed distribution is less than 12m/s.

Suppose the relationship between Y and X is given by: Y = 4 + 3.2*X + error  By how much does the expected value of Y change if X increases by 1.29 units? (Round your answer to two decimal places: ex: 123.45)

You are conducting a study to see if the probability of a true negative on a test for a certain cancer is significantly different from 63%. With H, :p 63% you obtain a test statistic of z = - 1.222. Find the p-value accurate to 4 decimal places. p-value =

Workers' Distractions. A recent study showed that the modern working person experiences an average of 2.1 hours per day of distractions (phone calls, e- mails, impromptu visits, etc.). A random sample of 16 workers for a large corporation found that these workers were distracted an average of 1.8 hours per day and the standard deviation was 20 minutes. Estimate the true mean population distraction time with 90% confidence, and compare your answer to the results of the study. (1.654,1.946). Mean of the study is outside or "different" from the result of our study.