This exercise has two parts. Please respond to both. The calculation formulas and procedures for both parts are included in the mini-lecture titled “Rates and Percentages” Part One The Mayors of two small towns that adjoin a large community have engaged in an argument on which of their cities is safer, with respect to its crime rate. City A has a residential population of 123,000 people and reported 185 violent crimes. City B has a residential population of 84,000 people and reported 135 violent crimes. Answer the following questions. 1. What is the reported crime rate per 100,000 residents in both cities? 2. Based on your responses to the previous question, which city has the lowest crime rate? Part Two Last year. 135 vehicles were reported stolen in the City of Bigton. This year, 185 vehicles were reported stolen in the City of Bigton. What is the percent change in reported stolen vehicles from last year to this year in the City of Bigton?
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1. The waiting times (in minutes) of a random sample of 21 people at a bank have a sample standard deviation of 3.5 minutes. Construct a confidence interval for the population variance sigma squared and the population standard deviation sigma. Use a 99 % level of confidence. Assume the sample is from a normally distributed population. What is the confidence interval for the population variance sigma squared?
2.You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 55 home theater systems has a mean price of $113.00. Assume the population standard deviation is $15.20.
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The Table in my homework question below is completely wrong. I am not sure where I went wrong in my calculations but coud you rework this question and answer the parts below??
Here are earnings per share for two companies by quarter from the first quarter of 2009 through the second quarter of 2012. Forecast earnings per share for the rest of 2012 and 2013. Use exponential smoothing to forecast the third period of 2012, and the time series decomposition method to forecast the last two quarters of 2012 and all four quarters of 2013. (It is much easier to solve this problem on a computer spreadsheet so you can see what is happening.) |
EARNINGS PER SHARE |
||||
QUARTER | COMPANY A | COMPANY B | ||
2009 | I | $ 1.68 | $ 0.21 | |
II | 2.36 | 0.23 | ||
III | 1.21 | 0.21 | ||
IV | 1.30 | 0.35 | ||
2010 | I | 1.66 | 0.20 | |
II | 2.08 | 0.36 | ||
III | 1.31 | 0.37 | ||
IV | 0.34 | 0.48 | ||
2011 | I | 0.34 | 0.35 | |
II | –0.19 | (loss) | 0.49 | |
III | –0.87 | (loss) | 0.51 | |
IV | 0.24 | 0.52 | ||
2012 | I | –1.65 | (loss) | 0.31 |
II | 0.37 | 0.52 | ||
a. |
For the exponential smoothing method, choose the first quarter of 2009 as the beginning forecast. Make two forecasts: one with α = 0.20 and one with α = 0.30. (Negative values should be indicated by a minus sign. Round your answers to 3 decimal places.)
|
In: Math
In: Math
Assume that we want to construct a confidence interval. Do one of the following, as appropriate: (a) find the critical value tα/2, (b) find the critical value zα/2, or (c) state that neither the normal distribution nor the t distribution applies. Here are summary statistics for randomly selected weights of newborn girls: n=151 x = 29.1 hg S =7.6 hg The confidence level is 95%. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. t Subscript alpha divided by 2 tα/2 = (Round to two decimal places as needed.) B. z Subscript alpha divided by 2 zα/2 = (Round to two decimal places as needed.) C. Neither the normal distribution nor the t distribution applies.
In: Math
Pls answer all three parts for UPVOTE
Data set:
Students Outside US
Stu ID | Age | GPA | Hrs spend on sch wrk |
3 | 48 | 4.00 | 7 |
6 | 47 | 2.79 | 14 |
9 | 45 | 3.48 | 5 |
12 | 19 | 4.00 | 30 |
15 | 24 | 3.10 | 10 |
18 | 34 | 3.24 | 2 |
21 | 44 | 36.00 | 6 |
24 | 19 | 2.85 | 7 |
27 | 19 | 2.80 | 10 |
30 | 27 | 3.40 | 8 |
33 | 28 | 2.90 | 16 |
36 | 27 | 3.40 | 8 |
39 | 28 | 2.90 | 16 |
42 | 21 | 2.9 | 4 |
45 | 20 | 2.50 | 6 |
48 | 23 | 3.3 | 18 |
51 | 41 | 3.80 | 7 |
54 | 21 | 2.60 | 26 |
57 | 39 | 5 | |
60 | 18 | 3.10 | 12 |
63 | 28 | 3.70 | 20 |
66 | 35 | 8 | |
69 | 37 | 2.80 | 6 |
72 | 21 | N/A | 21 |
75 | 20 | 3.00 | 4 |
78 | 30 | 3.50 | 6 |
81 | 21 | 3.1 | 4 |
84 | 21 | 3.20 | 3 |
87 | 37 | 2.86 | 3 |
90 | 19 | 3.30 | 12 |
Students in US:
Stu ID | Age | GPA | Hrs spend on sch wrk |
175 | 20 | 3.20 | 10 |
178 | 20 | 2.40 | 12 |
181 | 17 | 3.98 | 6 |
184 | 20 | 3.00 | 15 |
187 | 27 | 2.20 | 6 |
190 | 19 | 3.00 | 7 |
193 | 3.10 | 15 | |
196 | 20 | 6 | |
199 | 43 | 3.67 | 12 |
202 | 44 | 3.80 | 10 |
205 | 26 | 3.80 | 4 |
208 | 25 | 2.50 | 5 |
211 | 21 | 3.72 | 10 |
214 | 29 | 2.54 | 4 |
217 | 33 | 3.85 | 21 |
220 | 18 | 3.00 | 5 |
223 | 21 | 3.00 | 6 |
226 | 19 | 3.00 | 5 |
229 | 26 | 3.00 | 4 |
232 | 19 | 2.81 | 4 |
235 | 19 | 3.00 | 6 |
238 | 28 | 3.50 | 10 |
241 | 19 | 4.00 | 12 |
244 | 20 | 3.20 | 2 |
247 | 21 | 9 | |
250 | 20 | 2.51 | 3 |
253 | 20 | 3.20 | 5 |
256 | 23 | 2 | |
259 | 20 | 3.10 | 27 |
262 | 26 | 2.50 | 8 |
a) Construct a 5-number summary and boxplot using the variable “hours spent on school work at home” for both groups.
b) Compare the means for both groups to answer the research questions in the first paragraph. Which group has a higher mean GPA? Which group spends more time on their homework? What conclusions can you draw about students who were born in the USA and those who were born outside the USA based on this analysis?
c) Identify any outliers in both groups
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Justin is interested in buying a digital phone. He visited 9 stores at random and recorded the price of the particular phone he wants. The sample of prices had a mean of 337.63 and a standard deviation of 30.04.
(a) What t-score should be used as the multiplier for a 95%
confidence interval for the mean, ?, of the distribution?
t =
(b) Calculate a 95% confidence interval for the mean price of
this model of digital phone:
(Enter the smaller value in the left answer box.)
___ to ___
In: Math
The Pew Research Center Internet Project conducted a survey of 657 Internet users. This survey provided a variety of statistics on them. If required, round your answers to four decimal places.
(a) The sample survey showed that 90% of respondents said the Internet has been a good thing for them personally. Develop a 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally.
___ to ___
(b) The sample survey showed that 67% of Internet users said the Internet has generally strengthened their relationship with family and friends. Develop a 95% confidence interval for the proportion of respondents who say the Internet has strengthened their relationship with family and friends.
___ to ___
(c) Fifty-six percent of Internet users have seen an online group come together to help a person or community solve a problem, whereas only 25% have left an online group because of unpleasant interaction. Develop a 95% confidence interval for the proportion of Internet users who say online groups have helped solve a problem.
__ to __
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The types of browse favored by deer are shown in the following table. Using binoculars, volunteers observed the feeding habits of a random sample of 320 deer.
Type of Browse | Plant Composition in Study Area |
Observed Number of Deer Feeding on This Plant |
Sage brush | 32% | 105 |
Rabbit brush | 38.7% | 125 |
Salt brush | 12% | 46 |
Service berry | 9.3% | 22 |
Other | 8% | 22 |
Use a 5% level of significance to test the claim that the natural distribution of browse fits the deer feeding pattern.
(a) What is the level of significance?
State the null and alternate hypotheses.
H0: The distributions are different.
H1: The distributions are the same.
H0: The distributions are the same.
H1: The distributions are the
same.
H0: The distributions are the same.
H1: The distributions are different.
H0: The distributions are different.
H1: The distributions are different.
(b) Find the value of the chi-square statistic for the sample.
(Round the expected frequencies to at least three decimal places.
Round the test statistic to three decimal places.)
Are all the expected frequencies greater than 5?
Yes
No
What sampling distribution will you use?
binomial
Student's t
uniform
normal
chi-square
What are the degrees of freedom?
(c) Estimate the P-value of the sample test statistic.
P-value > 0.100
0.050 < P-value < 0.100
0.025 < P-value < 0.050
0.010 < P-value < 0.025
0.005 < P-value < 0.010
P-value < 0.005
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis that the population fits the
specified distribution of categories?
Since the P-value > α, we fail to reject the null hypothesis.
Since the P-value > α, we reject the null hypothesis.
Since the P-value ≤ α, we reject the null hypothesis.
Since the P-value ≤ α, we fail to reject the null hypothesis.
(e) Interpret your conclusion in the context of the
application.
At the 5% level of significance, the evidence is sufficient to conclude that the natural distribution of browse does not fit the feeding pattern.
At the 5% level of significance, the evidence is insufficient to conclude that the natural distribution of browse does not fit the feeding pattern.
In: Math
Historically, the time needed for college students to complete their degree follows a normal distribution with a mean of 4 years and a standard deviation of 1.2 years. You wish to see if the mean time m has changed in recent years, so you collect information from 5 recent college graduates.
4.25 4 3.75 4.5 5
Is there evidence that the mean time is different from 4 years?
a. Check the needed conditions for both the test statistic and confidence interval. (Do not do a stemplot.)
b. State Ho and Ha
c. Calculate the test statistic (if applicable – state the degrees of freedom)
d. Find the p-value
e. What is the conclusion for this problem? Do you reject Ho?
f. Calculate the 98% confidence interval.
g. Interpret this confidence interval
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QUESTION 2
Part 1
In a survey of 800 college students in the United States, 608 indicated that they believe that a student or faculty member on campus who uses language considered racist, sexist, homophobic, or offensive should be subject to disciplinary action. Assuming that the sample is representative of college students in the United States, construct a 95% confidence interval for the proportion of college students who have this belief. (Use a table or technology. Round your answers to three decimal places.)
(___,___)
PART B
Interpret the interval. Chose 1
A. We are 95% confident that the mean number of U.S. college students who believe that a student or faculty member on campus who uses language considered racist, sexist, homophobic, or offensive should be subject to disciplinary action falls within this interval.
B. There is a 95% chance that the true proportion of U.S. college students who believe that a student or faculty member on campus who uses language considered racist, sexist, homophobic, or offensive should be subject to disciplinary action falls within this interval.
C. There is a 95% chance that the true proportion of U.S. college students who believe that a student or faculty member on campus who uses language considered racist, sexist, homophobic, or offensive should be subject to disciplinary action falls directly in the middle of this interval.
D. We are 95% confident that the true proportion of U.S. college students who believe that a student or faculty member on campus who uses language considered racist, sexist, homophobic, or offensive should be subject to disciplinary action falls within this interval.
E. We are 95% confident that the true proportion of U.S. college students who believe that a student or faculty member on campus who uses language considered racist, sexist, homophobic, or offensive should be subject to disciplinary action falls directly in the middle of this interval.
In: Math
Cultivated amaranth grains (Amaranthus caudatus) come in three colors: black, brown, and pale. Geneticists asked whether this phenotype could result from a dominant epistatic control. Crossing black-seeded and pale-seeded A. caudatus populations (homozygous lines) gave the following counts of black, brown, and pale seeds in the second generation (F2).
Seed coat color |
Black |
Brown |
Pale |
Seed count |
344 |
82 |
31 |
In genetics, dominant epistasis should lead to 12/16 of all such seeds being black, 3/16 Brown, and 1/16 pale. Do the experimental data above support the conclusion that seed coat color in cultivated amaranth grains is due to dominant epistasis?
Please find the Ho, Ha, test statistic, df, exact probability of the test statistic, and conclusion relative to the hypothesis. Is this a 1-tailed or 2-tailed test? Use Excel functions and math calculations, please.
I think you Chi-Squared for this, but I am not sure.
In: Math
Why will a design for a Graeco-Latin square of size 3 not work well?
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Pick a profession, identify events that might affect the variation ( you must include all three ; cyclical, seasonal and irregular ) of the secular trend. List at least three ways that this information could be beneficial to the industry.
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To test whether the mean time needed to mix a batch of material is the same for machines produced by three manufacturers, the Jacobs Chemical Company obtained the following data on the time (in minutes) needed to mix the material.
Manufacturer |
||||
1 | 2 | 3 | ||
21 | 27 | 24 | ||
27 | 24 | 18 | ||
21 | 30 | 24 | ||
18 | 27 | 18 |
Sum of Squares, Treatment | |
Sum of Squares, Error | |
Mean Squares, Treatment | |
Mean Squares, Error |
In: Math