In an article in the Journal of Advertising, Weinberger and Spotts compare the use of humor in television ads in the United States and in the United Kingdom. Suppose that independent random samples of television ads are taken in the two countries. A random sample of 400 television ads in the United Kingdom reveals that 145 use humor, while a random sample of 500 television ads in the United States reveals that 124 use humor. (a) Set up the null and alternative hypotheses needed to determine whether the proportion of ads using humor in the United Kingdom differs from the proportion of ads using humor in the United States. H0: p1 − p2 0 versus Ha: p1 − p2 0. (b) Test the hypotheses you set up in part a by using critical values and by setting α equal to .10, .05, .01, and .001. How much evidence is there that the proportions of U.K. and U.S. ads using humor are different? (Round the proportion values to 3 decimal places. Round your answer to 2 decimal places.) z H0 at each value of α; evidence. (c) Set up the hypotheses needed to attempt to establish that the difference between the proportions of U.K. and U.S. ads using humor is more than .05 (five percentage points). Test these hypotheses by using a p-value and by setting α equal to .10, .05, .01, and .001. How much evidence is there that the difference between the proportions exceeds .05? (Round the proportion values to 3 decimal places. Round your z value to 2 decimal places and p-value to 4 decimal places.) z p-value H0 at each value of α = .10 and α = .05; evidence. (d) Calculate a 95 percent confidence interval for the difference between the proportion of U.K. ads using humor and the proportion of U.S. ads using humor. Interpret this interval. Can we be 95 percent confident that the proportion of U.K. ads using humor is greater than the proportion of U.S. ads using humor? (Round the proportion values to 3 decimal places. Round your answers to 4 decimal places.) 95% of Confidence Interval [ , ] the entire interval is above zero.

In an article in the Journal of Advertising, Weinberger and Spotts compare the use of humor in television ads in the United States and in the United Kingdom. Suppose that independent random samples of television ads are taken in the two countries. A random sample of 400 television ads in the United Kingdom reveals that 142 use humor, while a random sample of 500 television ads in the United States reveals that 126 use humor. (a) Set up the null and alternative hypotheses needed to determine whether the proportion of ads using humor in the United Kingdom differs from the proportion of ads using humor in the United States. H0: p1 − p2 0 versus Ha: p1 − p2 0. (b) Test the hypotheses you set up in part a by using critical values and by setting α equal to .10, .05, .01, and .001. How much evidence is there that the proportions of U.K. and U.S. ads using humor are different? (Round the proportion values to 3 decimal places. Round your answer to 2 decimal places.) z H0 at each value of α; evidence. (c) Set up the hypotheses needed to attempt to establish that the difference between the proportions of U.K. and U.S. ads using humor is more than .05 (five percentage points). Test these hypotheses by using a p-value and by setting α equal to .10, .05, .01, and .001. How much evidence is there that the difference between the proportions exceeds .05? (Round the proportion values to 3 decimal places. Round your z value to 2 decimal places and p-value to 4 decimal places.) z p-value H0 at each value of α = .10 and α = .05; evidence. (d) Calculate a 95 percent confidence interval for the difference between the proportion of U.K. ads using humor and the proportion of U.S. ads using humor. Interpret this interval. Can we be 95 percent confident that the proportion of U.K. ads using humor is greater than the proportion of U.S. ads using humor? (Round the proportion values to 3 decimal places. Round your answers to 4 decimal places.) 95% of Confidence Interval [ , ] the entire interval is above zero.

Problem 1: The average Saturday attendance at a movie theater is 974 people with a standard deviation of 54 people.

Part A: What is the probability that less than 900 people will attend this coming Saturday?

Part B: What is the probability of between 875 and 1075 people will attend this Saturday?

Part C: Eighty percent of Saturday attendances will be less than how many people?

Part D: The movie theater manager wants to determine a staffing level such that 98% of the time she can service the customers. How many customers should she set a staffing plan to serve?

2. In 2013, Corky St. Clair, an “off-off-off-off” Broadway producer/director moves back to Blaine Missouri and purchases an old movie theater in hopes of reviving the facility for future community theater productions. The purchase price of the property was $150,000. He paid $15,050 to replace the roof, paid $7,643 for wiring and plumbing to meet the city code, and hired local teenagers to sweep and clean the building totaling $320. Additional costs include: $13,000 to remodel backstage offices; $1,200 for advertising opening night; and $3,000 for the post-show party. What is the cost of building?

A home theater in a box is the easiest and cheapest way to provide surround sound for a home entertainment center. A sample of prices is shown here (Consumer Reports Buying Guide, 2004). The prices are for models with a DVD player and for models with a DVD player.

1. Compute the mean price for models with a DVD player and the mean price for models without a DVD player. What is the additional price paid to have a DVD player included in a home theater unit?
2. Compute the range, variance, and standard deviation for the two samples. What does this information tell you about the prices for models with and without a DVD player?

Suppose the JLFB movie production company has produced two movies, A and B, and distributes them to theaters with willingness to pay for the two movies, which is shown below,

Movie A Movie B

A. Smith Theater                         $135 $95

J. Schumpeter Theater                $85 $115

a. Which package will result in the largest profit for the JFLB movie production company?

b. Charge $85 for movie A and $95 for movie B.

c. Charge $135 for movie A and $115 for movie B.

d. Charge $100 for both movie A and B if they are bought together.

e. Charge $110 for both movie A and B if they are bought together.

as a marketing director for an hotel, illustrate with the use of a diagram how you can apply the service marketing mix to the hotel

1.a. In a survey carried out at a famous water park, 28 children out of a random sample of 80 said that they used the water slide regularly. Find a 95 % confidence interval for the true proportion of all children at the water park who uses the water slide regularly. [4]

b. The owner of the water park found that 45 children out of a random sample of 100 said that they used the pool regularly. Find a 98% confidence interval for the true proportion of all children in the water park who uses the pool. [4]

1. Let us assume that there are two visitors, A and B, in an amusement park. The demand curve for the visitors facing the amusement park are as follows.

P_A= 5 – 2Q_A

P_B= 2.5 – 0.5Q_B

Marginal cost (MC) to serve each visitor is equal to $1.

a. If the amusement park decides to set the price using two-part tariff, given the demand curve

P=6 – 2.5Q and MC =$1, how much is the equilibrium P and Q

b. Calculate the maximum upfront fee the park could charge each visitor

In Avocado Park, 38% of the population is made up of immigrants. Consider a random sample of 78 residents of Avocado Park.

a.How many individuals must reside in Avocado Park to consider the selection of these individuals to be independent?

b.What is the probability that more than 30 of the residents in the sample are immigrants?

c. How many immigrants should be EXPECTED to be in the sample?

d.Peacoat Fashions currently employs 78 Avocado Park residents. If there are fewer than 20 employees that are immigrants, does that suggest that immigrants are less likely to be hired at Peacoat Fashions?