Crime concerns in China. A 2013 poll found that 24.2% of Chinese adults see crime as a very big problem, and the standard error for this estimate, which can reasonably be modeled using a normal distribution, is SE = 1.3%. Suppose an issue will get special attention from the Chinese government if more than 1-in-5 Chinese adults express concern on an issue.

1. Choose words from the dropdown choices to construct hypotheses regarding whether or not crime should receive special attention by the Chinese government according to the 1-in-5 guideline. Before making your choices consider the appropriateness of using a one-sided or two-sided test for this exercise. That is, for this decision process, would we care about one or both directions?

H0:H0: The proportion of adults in China who see crime as a very big problem is  ? more than not more than different from not different from less than not less than   ? 20% 24.2% 1.3% . The observed difference  ? is is not  due to chance.

HA:HA: The proportion of adults in China who see crime as a very big problem is  ? more than not more than different from not different from less than not less than   ? 20% 24.2% 1.3% . The observed difference  ? is is not  due to chance.

2. Calculate a z-score using the observed percentage and the two model parameters. Round to four decimal places. z =

3. Use the normal model to calculate a p-value. Round to four decimal places. p =

4. Based on your p-value, should crime receive special attention from the Chinese government?

? Yes No  because we  ? should should not  reject the null hypothesis.

Clint Barton is a self-employed consultant who runs his company, Hawkeye Solutions, out of a house that he owns that he does not live in. He uses the main floor for his work and rents out the second floor to a tenant. Since he uses a fair amount of computing in his work, he wants to make sure that he writes off a representative portion of the electrical bill for the house against the business. In the past, he estimated that 70% of the electricity in the house goes towards the business.

Since the Canada Revenue Agency might want to see documentation about his expenses, Barton wants to sample some of the power use to provide support for his case. Since the building has central heating and cooling, these electrical costs are shared evenly as utilities and Barton has already accounted for them separately.

Barton connected monitors to lines going to each floor for non-utility power. Because he has had disagreements about proper expenses in the past, he wants to provide strong evidence to the Canada Revenue Agency to support his claim that he uses 70% of the non-utility power for the business.

Design a test for Barton where he will record the values of the power going upstairs and to the main floor every day for a month (30 days). And answer the following questions (use complete sentences and exact equations where possible):

Barton does the test as you outline above. He finds over the course of the sample that the business used $320 of power and the upstairs used $80.

f) What is the value of the test statistic in this case?

g) What is the decision of the test in this case? Explain your reasoning.

h) What does this mean for Barton’s attempt to get sufficient documentation?

A month later, Barton does a test with a sample of 2 months (60 days). He finds over the course of the sample that the business used $720 of power and the upstairs used $180.

i) What is the value of the test statistic in this case?

j) What is the decision of the test in this case? Explain your reasoning.  

k) What does this mean for Barton’s attempt to get sufficient documentation?  

t-test data,

Calculate the pooled variance for this dataset.

Group 1= 9.7, 9.5, 9.2, 8.5, 10.9, 9.8, 8.7, 9.8, 7.9, 9.0, 10.5, 8.9, 10.0, 8.9, 6.8, 8.2, 9.3, 10.5, 8.5, 9.4

Group 2= 8.1, 7.8, 7.6, 8.1, 9.9, 8.6, 8.8, 9.1, 10.4, 9.1, 6.9, 7.3, 6.7, 5.5, 9.6, 7.8, 8.7, 9.5, 7.8, 8.2

An ANOVA test was conducted in R with the numbers...

monkey group 1= 9.7, 9.2, 9.5, 9.5, 10.9, 9.8, 8.7, 7.9, 9.8, 9, 10.5, 8.9, 10, 8.9, 6.8, 9.3, 8.2, 8.5, 9.4, 10.5

Monkey group 2= 8.1, 7.8, 7.6, 9.9, 8.1, 9.1, 8.8, 10.4, 8.6, 6.9, 9.1, 6.7, 7.3, 7.8, 9.6, 8.7, 5.5, 9.5, 8.2, 7.8

What value in the ANOVA output would be identical to the pooled variance found from the t-test data?

Show how to answer this in EXCEL ONLY, NO megastat or minitab, etc.. Please highlight what data tools or formulas were used.

A survey investigated the public’s attitude toward the federal deficit. Each sampled citizen was classified as to whether he or she felt the government should reduce the deficit or increase the deficit, or if the individual had no opinion. The sample results of the study by gender are reported to below.

At the .05 significance level, is it reasonable to conclude that gender is independent of a person’s position on the deficit?

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 60 home theater systems has a mean price of $111.00. Assume the population standard deviation is $16.20.

The 90% confidence interval is(--,--)

The 99% confidence interval is(--,--)

Which interval is wider?

. You measure 36 textbooks' weights, and find they have a mean weight of 30 ounces. Assume the population standard deviation is 10 ounces. Based on this, construct a 90% confidence interval for the true population mean textbook weight.

Give your answers as decimals, to two places

Assume that a sample is used to estimate a population proportion p. Find the 80% confidence interval for a sample of size 155 with 138 successes. Enter your answer using decimals (not percents) accurate to three decimal places.

You measure 48 turtles' weights, and find they have a mean weight of 57 ounces. Assume the population standard deviation is 13.8 ounces. Based on this, determine the point estimate and margin of error for a 95% confidence interval for the true population mean turtle weight.

Give your answers as decimals, to two places

Ask Your Teacher Recent studies have shown that about 20% of American adults fit the medical definition of being obese. A large medical clinic would like to estimate what percentage of their patients are obese, so they take a random sample of 100 patients and find that 16 are obese. Suppose that in truth, the same percentage holds for the patients of the medical clinic as for the general population, 20%. If the clinic took repeated random samples of 100 observations and found the sample proportion who were obese, into what interval should those sample proportions fall about 95% of the time? (Round your answers to two decimal places. Between _____ and _____

You wish to test the following claim at a significance level of α=0.02

      H0:μ=62.5
      H1:μ>62.5

You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size 64 with mean 66.5 and a standard deviation of 12.7.

What is the test statistic for this sample? (Report answer accurate to 3 decimal places.)
test statistic =

What is the p-value for this sample? (Report answer accurate to 4 decimal places.)
p-value =

The number of men and women among professors in Math, Physics, Chemistry, Linguistics, and English departments from a SRS of small colleges were counted, and the results are shown in the table below.

Dept. Math Physics Chemistry Linguistics English

Men 16 . 36 12 10 14

Women 2 . 5 . 4 . 2 . 8

Test the claim that the gender of a professor is independent of the department. Use the significance level α=0.025

(a) The test statistic is χ2=

(b) The critical value is χ2=

(c) Is there sufficient evidence to warrant the rejection of the claim that the gender of a professor is independent of the department? A. No B. Yes

In a sample of 169 trees, we found that a pear tree grow to average height of 32 feet and a sample standard deviation of 5 feet. The distribution is approximately normal. Find the 95% confidence interval for the mean population.

For a sample of eight​ bears, researchers measured the distances around the​ bears' chests and weighed the bears. Minitab was used to find that the value of the linear correlation coefficient is r=0.931 Using alphaα=​0.05, determine if there is a linear correlation between chest size and weight. What proportion of the variation in weight can be explained by the linear relationship between weight and chest​ size?

a. Is there a linear correlation between chest size and​ weight?

A.​Yes, because the absolute value of r exceeds the critical value of 0.707

B.​No, because the absolute value of r exceeds the critical value of 0.707.

C.​Yes, because r falls between the critical values of −0.707 and 0.707.

D. The answer cannot be determined from the given information.

b. What proportion of the variation in weight can be explained by the linear relationship between weight and chest​ size?

A company wants to build a plant but is considering the size. The table below shows their payoffs under different states of demand.

They can hire a consultant who can conduct a survey to evaluate demand. The consultant will report to the company whether demand is strong or weak. The probabilities are 0.60 and 0.40 for strong and weak survey results respectively. The conditional probabilities for demand given survey results are as follow:

P(Low/Strong) = 0.35; P(High/Strong) = 0.65;

P(Low/Weak) = 0.70; P(High/Weak) = 0.30;

a) Draw the decision tree for this problem including with and without a survey.

b) What is the best decision without a survey?

c) What is the decision strategy when survey is conducted?

d) What is the EVPI?

e) What is the EVII

A random variable X has density function f(x) = 4x ( 1 + x²)^-3 for x > 0.

Determine the mode of X.

A survey of 1060people who took trips revealed that 94 of them included a visit to a theme park. Based on those survey results, a management consultant claims that less than 11 % of trips include a theme park visit. Test this claim using the ?=0.01significance level.

(a) The test statistic is ___

(b) The P-value is ___

(c) The conclusion is  

A. There is sufficient evidence to support the claim that less than 11 % of trips include a theme park visit.
B. There is not sufficient evidence to support the claim that less than 11 % of trips include a theme park visit.

Independent random samples, each containing 90 observations, were selected from two populations. The samples from populations 1 and 2 produced 36 and 26 successes, respectively.
Test ?0:(?1−?2)=0against ??:(?1−?2)>0 Use ?=0.1

(a) The test statistic is ___

(b) The P-value is ___

(c) The final conclusion is
A. There is not sufficient evidence to reject the null hypothesis that (?1−?2)=0
B. We can reject the null hypothesis that (?1−?2)=0 and conclude that (?1−?2)>0

A random sample of 1500 residential telephones in Phoenix found that 385 of the numbers were unlisted. A random sample in the same year of 1200 telephones in Scottsdale found that 311 were unlisted.

Round your answers to four decimal places (e.g. 98.7654).

(a) Calculate a 95% two-sided confidence interval on the difference in the proportions of unlisted numbers between the two cities.
Enter your answer; 95% confidence interval, lower bound ≤p1-p2≤ Enter your answer; 95% confidence interval, upper bound

(b) Is there a significant difference between the two proportions at α = 0.05? Choose your answer in accordance to the item b) of the question statement                                                          No.Yes.

(c) Calculate a 90% two-sided confidence interval on the difference in the proportions of unlisted numbers between the two cities.
Enter your answer; 90% confidence interval, lower bound ≤p1-p2≤ Enter your answer; 90% confidence interval, upper bound