Chat with us, powered by LiveChat Homework 4 Hypothesis ANOVA and tTest – Statistics Assignment | Abc Paper
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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 **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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