4. The only thing that changes in Dullsville is the price of a stay at the Dullsville Inn. You've collected the following data on the rates charged (for a suite with 2 queen-sized beds and 'free' continental breakfast) and the number of rooms occupied. The Inn has 100 suites, and at no time were potential visitors turned away due to no vacancy. Use this data to estimate a 'constant elasticity' demand function. Estimate the price elasticity of demand.                                                                                                                     

                Observation          Rate per night       Quantity (rooms rented)                   

                1              $70         40                          

                2              $65         50                          

                3              $80         30                          

                4              $52         62                          

                5              $92         31                          

                6              $64         41                          

                7              $43         78                          

                8              $74         35                          

                9              $83         33                          

                10           $54         52                          

                11           $87         30                          

                12           $84         28                          

                13           $68         40                          

                14           $43         69                          

                15           $48         53                          

                16           $78         34                          

                17           $72         48                          

                18           $58         53                          

                19           $56         59      

1) You measure 42 textbooks' weights, and find they have a mean weight of 64 ounces. Assume the population standard deviation is 9.6 ounces. Based on this, construct a 99% confidence interval for the true population mean textbook weight. Give your answers as decimals, to two places.

__< μ < __

2) Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 70 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 17.2 and a standard deviation of 3.9. What is the 80% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Enter your answers accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).
__ < μ < __

3) 44% of 84 delegates in a policitcal convention favored changing the rules to restrict the number of potential candidates. What is the 80% confidence interval for the population proportion.
Give your answers as decimals, to two places.

__ < p < __

1. NPV: Project L costs $55,000, its expected cash inflows are $12,000 per year for 7 years, and its WACC is 10%. What is the project's payback? Round your answer to two decimal places.

2.IRR: Project L costs $72,976.35, its expected cash inflows are $14,000 per year for 11 years, and its WACC is 12%. What is the project's IRR? Round your answer to two decimal places.

3. PAYBACK PERIOD: Project L cOSTS $60,000, its expected cash inflows are $13,000 per year for 10 years, and its WACC is 11%. What is the project's payback? Round your answer to two decimal places _______ Years

4. DISCOUNTED PAYBACK: Project L costs $45,000, its expected cash inflows are $11,000 per year for 8 years, and its WACC is 8%. What is the project's discounted payback. Round your answer to two decimal places. _______ years

1. Explain under what conditions a researcher would use a One-Way Between-Subjects ANOVA.

2. A psychologist at a private mental hospital was asked to determine whether there was a clear difference in the length of stay for patients with different categories of diagnosis. Looking at the last four patients in each of the three major categories, the results (in terms of weeks of stay) were as follows:

Diagnosis Category

Using alpha = 0.05, is there a significant difference in length of stay among diagnosis categories? Show all five steps of hypothesis testing.

3. Compute the means and standard deviations for each group.

Regular expression translation

1. [ac7]b[bc]
   Translation: Match any 1 character in a set that is either a, c, or 7, then a b, then any 1 character that is in a set that is either b or c.
   Match example: abc
   
2. ^…$
   
3. \<the\>
   
4. ^[A-Z]..$
   
5. ^[A-Z][ a-z]*3[0-5]
   
6. [a-z]*\.$
   
7. ^ *[A-Z][a-z][a-z]$
   
8. ^[A-Za-z]*[^,][A-Za-z]*$
   
9. [[:digit:]]{1,3}\.[[:digit:]]{1,3}\.[[:digit:]]{1,3}\.[[:digit:]]{1,3}
   
10. .*\([0-9][.][0-9]\)
    
11. [0-9]{3}-[0-9]{4}

Consider the two variables service quality and customer satisfaction. Service quality is independent variable and customer satisfaction is dependent variable. Perform regression analysis using SPSS and explain the results.

PLEASE USE SPSS ONLY AND PASTE THE OUTPUT OF SPSS. DO NOT USE EXCEL OR ANY OTHER SOFTWARE. SPSS ONLY PLEASE.

Calculation of individual costs and WACCLang Enterprises is interested in measuring its overall cost of capital. Current investigation has gathered the following data. The firm is in the 27% tax bracket.

Debt The firm can raise debt by selling $1,000​-par-value, 8​% coupon interest​ rate,17-year bonds on which annual interest payments will be made. To sell the​ issue, an average discount of

$50 per bond would have to be given. The firm also must pay flotation costs of $20 per bond.

Preferred stock The firm can sell 7% preferred stock at its $100-per-share par value. The cost of issuing and selling the preferred stock is expected to be $4 per share. Preferred stock can be sold under these terms.

Common stock The​ firm's common stock is currently selling for $55 per share. The firm expects to pay cash dividends of $7.5 per share next year. The​ firm's dividends have been growing at an annual rate of 8%, and this growth is expected to continue into the future. The stock must be underpriced by $44 per​share, and flotation costs are expected to amount to $6 per share. The firm can sell new common stock under these terms.

Retained earnings When measuring this​ cost, the firm does not concern itself with the tax bracket or brokerage fees of owners. It expects to have available $130,000 of retained earnings in the coming​ year; once these retained earnings are​ exhausted, the firm will use new common stock as the form of common stock equity financing.

a.  Calculate the​ after-tax cost of debt.

b.  Calculate the cost of preferred stock.

c.  Calculate the cost of common stock.

d.  Calculate the​ firm's weighted average cost of capital using the capital structure weights shown in the following​ table… (Round answer to the nearest​ 0.01%)

Question 5

An urban planner is researching commute times in the San Francisco Bay Area to find out if commute times have increased. In which of the following situations could the urban planner use a hypothesis test for a population mean? Check all that apply.

1. The urban planner asks a simple random sample of 110 commuters in the San Francisco Bay Area if they believe their commute time has increased in the past year. The urban planner will compute the proportion of commuters who believe their commute time has increased in the past year.
2. The urban planner collects travel times from a random sample of 125 commuters in the San Francisco Bay Area. A traffic study from last year claimed that the average commute time in the San Francisco Bay Area is 45 minutes. The urban planner will see if there is evidence the average commute time is greater than 45 minutes.
3. The urban planner asks a random sample of 100 commuters in the San Francisco Bay Area to record travel times on a Tuesday morning. One year later, the urban planner asks the same 100 commuters to record travel times on a Tuesday morning. The urban planner will see if the difference in commute times shows an increase.

Question 6

The Food and Drug Administration (FDA) is a U.S. government agency that regulates (you guessed it) food and drugs for consumer safety. One thing the FDA regulates is the allowable insect parts in various foods. You may be surprised to know that much of the processed food we eat contains insect parts. An example is flour. When wheat is ground into flour, insects that were in the wheat are ground up as well.

The mean number of insect parts allowed in 100 grams (about 3 ounces) of wheat flour is 75. If the FDA finds more than this number, they conduct further tests to determine if the flour is too contaminated by insect parts to be fit for human consumption.

The null hypothesis is that the mean number of insect parts per 100 grams is 75. The alternative hypothesis is that the mean number of insect parts per 100 grams is greater than 75.

Is the following a Type I error or a Type II error or neither?

The test fails to show that the mean number of insect parts is greater than 75 per 100 grams when it is.

1. Type I error
2. Type II error
3. Neither

Question 7

Child Health and Development Studies (CHDS) has been collecting data about expectant mothers in Oakland, CA since 1959. One of the measurements taken by CHDS is the age of first time expectant mothers. Suppose that CHDS finds the average age for a first time mother is 26 years old. Suppose also that, in 2015, a random sample of 50 expectant mothers have mean age of 26.5 years old, with a standard deviation of 1.9 years. At the 5% significance level, we conduct a one-sided T-test to see if the mean age in 2015 is significantly greater than 26 years old. Statistical software tells us that the p-value = 0.034.

Which of the following is the most appropriate conclusion?

1. There is a 3.4% chance that a random sample of 50 expectant mothers will have a mean age of 26.5 years old or greater if the mean age for a first time mother is 26 years old.
2. There is a 3.4% chance that mean age for all expectant mothers is 26 years old in 2015.
3. There is a 3.4% chance that mean age for all expectant mothers is 26.5 years old in 2015.
4. There is 3.4% chance that the population of expectant mothers will have a mean age of 26.5 years old or greater in 2015 if the mean age for all expectant mothers was 26 years old in 1959.

Question 8

Child Health and Development Studies (CHDS) has been collecting data about expectant mothers in Oakland, CA since 1959. One of the measurements taken by CHDS is the weight increase (in pounds) for expectant mothers in the second trimester. In a fictitious study, suppose that CHDS finds the average weight increase in the second trimester is 14 pounds. Suppose also that, in 2015, a random sample of 40 expectant mothers have mean weight increase of 16 pounds in the second trimester, with a standard deviation of 6 pounds. At the 5% significance level, we can conduct a one-sided T-test to see if the mean weight increase in 2015 is greater than 14 pounds. Statistical software tells us that the p-value = 0.021.

Which of the following is the most appropriate conclusion?

1. There is a 2.1% chance that a random sample of 40 expectant mothers will have a mean weight increase of 16 pounds or greater if the mean second trimester weight gain for all expectant mothers is 14 pounds.
2. There is a 2.1% chance that mean second trimester weight gain for all expectant mothers is 14 pounds in 2015.
3. There is a 2.1% chance that mean second trimester weight gain for all expectant mothers is 16 pounds in 2015.
4. There is 2.1% chance that the population of expectant mothers will have a mean weight increase of 16 pounds or greater in 2015 if the mean second trimester weight gain for all expectant mothers was 14 pounds in 1959.

Question 9

A researcher conducts an experiment on human memory and recruits 15 people to participate in her study. She performs the experiment and analyzes the results. She uses a t-test for a mean and obtains a p-value of 0.17.

Which of the following is a reasonable interpretation of her results?

1. This suggests that her experimental treatment has no effect on memory.
2. If there is a treatment effect, the sample size was too small to detect it.
3. She should reject the null hypothesis.
4. There is evidence of a small effect on memory by her experimental treatment.

Question 10

A criminal investigator conducts a study on the accuracy of fingerprint matching and recruits a random sample of 35 people to participate. Since this is a random sample of people, we don’t expect the fingerprints to match the comparison print. In the general population, a score of 80 indicates no match. Scores greater than 80 indicate a match. If the mean score suggests a match, then the fingerprint matching criteria are not accurate.

The null hypothesis is that the mean match score is 80. The alternative hypothesis is that the mean match score is greater than 80.

The criminal investigator chooses a 5% level of significance. She performs the experiment and analyzes the results. She uses a t-test for a mean and obtains a p-value of 0.04.

Which of the following is a reasonable interpretation of her results?

1. This suggests that there is evidence that the mean match score is greater than 80. This suggests that the fingerprint matching criteria are not accurate.
2. If there is a treatment effect, the sample size was too small to detect it. This suggests that we need a larger sample to determine if the fingerprint matching criteria are not accurate.
3. She cannot reject the null hypothesis. This suggests that the fingerprint matching criteria could be accurate.
4. This suggests that there is evidence that the mean match score is equal to 80. This suggests that the fingerprint matching criteria is accurate.

Question 11

A group of 42 college students from a certain liberal arts college were randomly sampled and asked about the number of alcoholic drinks they have in a typical week. The purpose of this study was to compare the drinking habits of the students at the college to the drinking habits of college students in general. In particular, the dean of students, who initiated this study, would like to check whether the mean number of alcoholic drinks that students at his college in a typical week differs from the mean of U.S. college students in general, which is estimated to be 4.73.

The group of 42 students in the study reported an average of 5.31 drinks per with a standard deviation of 3.93 drinks.

Find the p-value for the hypothesis test.

The p-value should be rounded to 4-decimal places.

Question 12

Commute times in the U.S. are heavily skewed to the right. We select a random sample of 240 people from the 2000 U.S. Census who reported a non-zero commute time.

In this sample the mean commute time is 28.9 minutes with a standard deviation of 19.0 minutes. Can we conclude from this data that the mean commute time in the U.S. is less than half an hour? Conduct a hypothesis test at the 5% level of significance.

What is the p-value for this hypothesis test?

Your answer should be rounded to 4 decimal places.

Match the margin of error for an 80% confidence interval to estimate the population mean with sigma σ equals=50

with its corresponding sample sizes.

Question 5 options:

a.) 8.07. 1.) n= 34

b.) 9.45 2.) n= 46

c.) 10.99 3.) n= 63

C - programming problem:

Let T be a sequence of an even length 2k, for some non-negative integer k. If its prefix of k symbols (or, the first k symbols in T) are the same as its suffix of k symbols (respectively, the last k symbols in T), then T is called a tandem sequence. For example, T[8] = 31103110 is a tandem sequence. Given a sequence S, if T is a subsequence of S and T is tandem, then T is called a tandem subsequence of S. One can then similarly define what a longest tandem subsequence (LTS for short) of S is.

Write a program that reads in two sequences with max length: 10,000, which computes, and prints the longest-tandem subsequence or LTS, and its length.

Sample program run:

Enter a sequence: 3110283025318292818318143

# an LTS (length = 10) for the sequence is:

1283312833