**Hi, please help me with the 3 questions below. I have the answers posted for 2 questions, can you review and correct if they are wrong and for the 3rd question I need you to work out the whole answer **please have all corrections done in excel and Q3 in excelsee attachment

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**Hi, please help me with the 3 questions below. I have the answers posted for 2 questions, can you

review and correct if they are wrong and for the 3rd question I need you to work out the whole answer

**

Question 1

Suppose a fast-food researcher is interested in determining if there is a difference between Denver

and Chicago in the average price of a comparable hamburger. There is some indication, based on

information published by Grub Hub, that the average price of a hamburger in Denver may be more

than it is in Chicago. Suppose further that the prices of hamburgers in any given city are

approximately normally distributed with a population standard deviation of $0.64. A random sample

of 15 different fast-food hamburger restaurants is taken in Denver and the average price of a

hamburger for these restaurants is $9.11. In addition, a random sample of 18 different fast-food

hamburger restaurants is taken in Chicago and the average price of a hamburger for these

restaurants is $8.62. Use techniques presented in this chapter to answer the researcher’s question.

Explain your results.

Answer

The research question is “if there is a difference between Denver and Chicago in the average price

of a comparable hamburger” so alternative hypothesis is

The null hypothesis is

Given information:

Let

The critical values are -1.96 and 1.96.

Rejection region:

If z < -1.96 or z > 1.96, reject H0

Test statistics will be

Here test is two tailed. So p-value of the test is

2P(z > 2.19) = 0.0285

Since p-value is less than 0.05 so we reject the null hypothesis.

That is at 5% level of significance, we can conclude that there is a difference between Denver and

Chicago in the average price of a comparable hamburger.

Question 2

Pipeline operators estimate that it costs between $75 and $500 in U.S. currency to replace each

seal, thus making the Clarkson longer-lasting valves more attractive. Tyco does business with

pipeline companies around the world. Suppose in an attempt to develop marketing materials, Tyco

marketers are interested in determining whether there is a significant difference in the cost of

replacing pipeline seals in different countries. Four countries-Canada, Colombia, Taiwan, and the

United States-are chosen for the study. Pipeline operators from equivalent operations are selected

from companies in each country. The operators keep a cost log of seal replacements. A random

sample of the data follows. Show the ANOVA output using EXCEL, discuss your table, report the

findings, and write your recommendation. Use these data to help Tyco determine whether there is a

difference in the cost of seal replacements in the various countries using Tukey’s HSD test. (a = .05)

Canada

Colombia

China

United States

$215

$355

$170

$230

205

280

190

190

245

300

235

225

270

330

195

220

290

360

205

215

260

340

180

245

225

300

190

230

Answer

Question 3 – I need you to work out this answer..

Refer back to your dataset used in Seminar 3 homework problems. Now, let’s assume that the

package claims that it contains 1.7 oz. of mustard seed. Some of the customers claim that there isn’t

enough mustard seed in the 1.7 oz. economy size; corporate management is worried that there may

be too much. Given the sample of 36, assuming that it is adequate, test the hypothesis that the

average amount of mustard seed in the package meets the 1.7 oz. standard with, say, 95%

confidence. Be sure to state the null and alternative hypothesis and which you support. Include your

Excel output. For extra assistance, reference section 9.3 and page 353 > USING THE COMPUTER

> Excel > “To perform a T test of a single mean”, etc. Is there anything you would want to note for

management?

Hints: This might help with the third problem in this week’s HW.

You will use the database from the problem from Seminar Three

homework. This particular explanation relates to Dataset #1. As you

recall, the data had a mean of 1.733056 and a standard error of .025,

with a 95% confidence interval of approximately 1.68-1.78 ounces, which

interval would account for approximately 95% of the means of all similar

samples over time.

Since Excel doesn’t do one sample T tests well, we’ll just trick it into

giving us what we want. Have students set up a second sample; call it

True Mean or whatever. Every sample in that column will be exactly 1.7 (see below).

Your null hypothesis is that the mean of variable 1 = the mean of variable 2; that is,

that the mean of variable 1 is not

significantly different from 1.7, or that differences are only

coincidental. (We can see that 1.733056 is different from 1.7… t he issue

is whether we can definitively say at a two tailed confidence of .05

that the means are significantly different. The alternative hypothesis is that the two

means are significantly different.

Once the data has been entered, you should have two columns. Column 1 is the

data from the dataset in Seminar 3, problem 1 (in this example, we use

Dataset #1); Column #2 contains 36 samples, all equal to 1.7. Go to Data

Analysis, then TTest Paired Two Sample for Means. Cursor through the

first column of data for the variable 1 range, and the second column for

the variable 2 range. If you include the labels, be sure to check the

Label box. The significance of .05 should be sufficient; then OK. The

output should resemble the bottom.

For the first problem in HW, use the t-stat formula in the text and calculate a t score by

hand. Use the Tutorial for help with ANOVA. You should use the data analysis tool for

the ANOVA problem.

…

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