Fowle marketing research, inc.. bases charges to a client on the assumption that telephone surveys can be completedwithin 15 minutes or less. if more time is required. a premium rate is charged. with a sample of 35 surveys, a populationstandard deviation of 4 minutes, and a level of significance of .01, the sample mean will be used to test the null hypothesish0: u s 15.a. what is your interpretation of the type ii error for this problem? what is its impact on the firm? b. what is the probability of making a type ii error when the actual mean time is u = 17 minutes? c. what is the probability of making a type ii error when the actual mean time is u = 18 minutes? d. sketch the general shape of the power curve for this test.

the answer is:

the correct option:

b.

to solve the problem, we need to remember how to operate to simplify exponential expressions using the logarithms.

we are given the equation:

then, using placing logarith with base 7 on both sides of the equation, we have:

so, rounding the obtained result to two decimal places (nearest hundreth), we have:

hence, we have that the correct option:

b.

have a nice day!

55.44 inches

step-by-step explanation:

part a. assuming a perfectly linear relationship, we can find the slope from the first two data points.

slope = m = (change in rainfall)/(change in years)

= (61.80 -62.33)/(2005 -2004) = -0.53/1 = -0.53

then the point-slope form of the equation of the line can be written as

y = m(x -h) +k . . . . . m = -0.53, (h, k) = (0, 62.33)

y = -0.53x +62.33 . . . x = years after 2004

part b. in 2017, x = 2017 -2004 = 13. then the predicted rainfall is

y = -0.53·13 +62.33 = 55.44 . . inches

the predicted rainfall in 2017 is 55.44 in.

step-by-step explanation:

its payment history because your score is mainly based on how consistent u are with your payment and if there on time. hope this : )