Write an essay:

Mary is the production manager of the potato chip division. She gets bonus equal to 10% of her base salary if the division meets its target profit. She realized that the preliminary profit figures for the division is $10,000 below the target profit. She is considering changing the estimated percentage completion with respect to materials and conversion in ending work in process inventory of the final processing department to increase the profit. Does Mary want the percentage completion to be increased or decreased? Briefly explain your reasoning. Do you think Mary should change the percentage completion? Why or why not?

One of your peers claims that boys do better in math classes than girls. Together you run two independent simple random samples and calculate the given summary statistics of the boys and the girls for comparable math classes. In Calculus, 15 boys had a mean percentage of 82.3 with standard deviation of 5.6 while 12 girls had a mean percentage of 81.2 with standard deviation of 6.7. What assumptions need to be made in order to determine the 90% confidence interval for the difference in the mean percentage scores for the boys in calculus and the girls in calculus? Supposing the assumption is true, calculate the interval.

In a study of government financial aid for college students, it becomes necessary to estimate the percentage of full time college students who earn a bachelors degree in four years or less. Find the sample size needed to estimate that percentage. Use a 0.05 margin of error and use a confidence level of 99%.

A. Assume that nothing is known about the percentage to be estimated

N =

B. Assume prior studies have shown that about 35% of full time students earn bachelors degrees in four years or less

C. Does the added knowledge in part b have much of an effect on the sample size?

The following partial information is available. Complete the table by filling in all the blanks. Each case is independent.

​(Round the contribution margin percentage to the nearest whole percent. Use parentheses or a minus sign for operating​ losses.)

Brief Exercise 13-6

Net income was $501,000 in 2015, $471,000 in 2016, and $519,000 in 2017. What is the percentage of change from (a) 2015 to 2016, and (b) from 2016 to 2017? Is the change an increase or a decrease? (If amount and percentage are a decrease show the numbers as negative, e.g. -55,000, -20% or (55,000), (20%). Round percentages to 0 decimal places, e.g. 12%.)

Limit of Liability (LOL) method- the amount that an insurer would pay for a particular loss if they were the ONLY insurer covering it.

Example:       

You buy two policies from Insurers A and B, respectively: FA_A = $20,000 and FA_B = $80,000. If a loss = $50,000 occurs and it is covered by both contracts, how do insurers pay for this loss?

1. Determine a percentage to split the loss:

1. The percentage of the loss should Insurer A pay for =
2. The percentage of the loss should Insurer B pay for =

2. Determine the amount payable by each insurer:

1. How much should Insurer A pay for the claim?
2. How much should Insurer B pay for the claim?

Based on data from a​ college, scores on a certain test are normally distributed with a mean of 1518 and a standard deviation of 324.

Standard score   Percent
-3.0   0.13
-2.5   0.62
-2   2.28
-1.5   6.68
-1   15.87
-0.9   18.41
-0.5   30.85
-0.1   46.02
0   50.00
0.10   53.98
0.5   69.15
0.9   81.59
1   84.13
1.5   93.32
2   97.72
2.5   99.38
3   99.87
3.5   99.98

Find the percentage of scores greater than

2166

Find the percentage of scores less than

1194

Find the percentage of scores between

870

and

1680.

Please show your work.

A. If a normal distribution of scores has a mean of 100 and a standard deviation of 10, what percentage of scores would lie below 70? a. 0.13%. b.2.15%. c. 2.28% d. 99.87%

B. What percentage of scores lie between 85 and 100 for a normal distribution with a mean of 100 and a standard deviation of 15? a. +15% b.-15% c.34.13% d. -34.13%

C What percentage of scores lie between 70 and 80 for a normal distribution with the mean of 100 and a standard deviation of 10? a.+27% b.-13% c.-7%. d. 2.15%.

In an article in Marketing Science, Silk and Berndt investigate the output of advertising agencies. They describe ad agency output by finding the shares of dollar billing volume coming from various media categories such as network television, spot television, newspapers, radio, and so forth. (a) Suppose that a random sample of 401 U.S. advertising agencies gives an average percentage share of billing volume from network television equal to 7.49%, and assume that σ equals 1.43 percent. Calculate a 95% confidence interval for the mean percentage share of billing volume from network television for the population of all U.S. advertising agencies. (Round your answers to 3 decimal places.) The 95% confidence interval is [ , ]. (b) Suppose that a random sample of 401 U.S. advertising agencies gives an average percentage share of billing volume from spot television commercials equal to 12.40%, and assume that σ equals 1.52 percent. Calculate a 95% confidence interval for the mean percentage share of billing volume from spot television commercials for the population of all U.S. advertising agencies. (Round your answers to 3 decimal places.) The 95% confidence interval is [ , ]. (c) Compare the confidence intervals in parts a and b. Does it appear that the mean percentage share of billing volume from spot television commercials for U.S. advertising agencies is greater than the mean percentage share of billing volume from network television? Explain. , confidence interval in (b) is totally the confidence interval in (a).

In an article in Marketing Science, Silk and Berndt investigate the output of advertising agencies. They describe ad agency output by finding the shares of dollar billing volume coming from various media categories such as network television, spot television, newspapers, radio, and so forth.

(a) Suppose that a random sample of 405 U.S. advertising agencies gives an average percentage share of billing volume from network television equal to 7.42%, and assume that σ equals 1.44 percent. Calculate a 95% confidence interval for the mean percentage share of billing volume from network television for the population of all U.S. advertising agencies. (Round your answers to 3 decimal places.)

The 95% confidence interval is            [, ].

(b) Suppose that a random sample of 405 U.S. advertising agencies gives an average percentage share of billing volume from spot television commercials equal to 12.40%, and assume that σ equals 1.56 percent. Calculate a 95% confidence interval for the mean percentage share of billing volume from spot television commercials for the population of all U.S. advertising agencies. (Round your answers to 3 decimal places.)

The 95% confidence interval is            [, ].

(c) Compare the confidence intervals in parts a and b. Does it appear that the mean percentage share of billing volume from spot television commercials for U.S. advertising agencies is greater than the mean percentage share of billing volume from network television? Explain.

(Click to select)YesNo , confidence interval in (b) is totally (Click to select)belowabove the confidence interval in (a).