The following data represent crime rates per 1000 population for a random sample of 46 Denver neighborhoods.† 63.2 36.3 26.2 53.2 65.3 32.0 65.0 66.3 68.9 35.2 25.1 32.5 54.0 42.4 77.5 123.2 66.3 92.7 56.9 77.1 27.5 69.2 73.8 71.5 58.5 67.2 78.6 33.2 74.9 45.1 132.1 104.7 63.2 59.6 75.7 39.2 69.9 87.5 56.0 154.2 85.5 77.5 84.7 24.2 37.5 41.1 (a) Use a calculator with mean and sample standard deviation keys to find the sample mean x and sample standard deviation s. (Round your answers to one decimal place.) x = crimes per 1000 people s = crimes per 1000 people (b) Let us say the preceding data are representative of the population crime rates in Denver neighborhoods. Compute an 80% confidence interval for μ, the population mean crime rate for all Denver neighborhoods. (Round your answers to one decimal place.) lower limit crimes per 1000 people upper limit crimes per 1000 people (c) Suppose you are advising the police department about police patrol assignments. One neighborhood has a crime rate of 62 crimes per 1000 population. Do you think that this rate is below the average population crime rate and that fewer patrols could safely be assigned to this neighborhood? Use the confidence interval to justify your answer. Yes. The confidence interval indicates that this crime rate is below the average population crime rate. Yes. The confidence interval indicates that this crime rate does not differ from the average population crime rate. No. The confidence interval indicates that this crime rate is below the average population crime rate. No. The confidence interval indicates that this crime rate does not differ from the average population crime rate. (d) Another neighborhood has a crime rate of 76 crimes per 1000 population. Does this crime rate seem to be higher than the population average? Would you recommend assigning more patrols to this neighborhood? Use the confidence interval to justify your answer. Yes. The confidence interval indicates that this crime rate does not differ from the average population crime rate. Yes. The confidence interval indicates that this crime rate is higher than the average population crime rate. No. The confidence interval indicates that this crime rate is higher than the average population crime rate. No. The confidence interval indicates that this crime rate does not differ from the average population crime rate. (e) Compute a 95% confidence interval for μ, the population mean crime rate for all Denver neighborhoods. (Round your answers to one decimal place.) lower limit crimes per 1000 people upper limit crimes per 1000 people (f) Suppose you are advising the police department about police patrol assignments. One neighborhood has a crime rate of 62 crimes per 1000 population. Do you think that this rate is below the average population crime rate and that fewer patrols could safely be assigned to this neighborhood? Use the confidence interval to justify your answer. Yes. The confidence interval indicates that this crime rate is below the average population crime rate. Yes. The confidence interval indicates that this crime rate does not differ from the average population crime rate. No. The confidence interval indicates that this crime rate is below the average population crime rate. No. The confidence interval indicates that this crime rate does not differ from the average population crime rate. (g) Another neighborhood has a crime rate of 76 crimes per 1000 population. Does this crime rate seem to be higher than the population average? Would you recommend assigning more patrols to this neighborhood? Use the confidence interval to justify your answer. Yes. The confidence interval indicates that this crime rate does not differ from the average population crime rate. Yes. The confidence interval indicates that this crime rate is higher than the average population crime rate. No. The confidence interval indicates that this crime rate is higher than the average population crime rate. No. The confidence interval indicates that this crime rate does not differ from the average population crime rate. (h) In previous problems, we assumed the x distribution was normal or approximately normal. Do we need to make such an assumption in this problem? Why or why not? Hint: Use the central limit theorem. Yes. According to the central limit theorem, when n ≥ 30, the x distribution is approximately normal. Yes. According to the central limit theorem, when n ≤ 30, the x distribution is approximately normal. No. According to the central limit theorem, when n ≥ 30, the x distribution is approximately normal. No. According to the central limit theorem, when n ≤ 30, the x distribution is approximately normal.

use DMAIC to improve a process in one or more of the organizations listed below:

Health-care facility

the NIH wants to prove that taking daily asprine reduce the chance of second heart attack with in five years. wiyhout taking the asprine it is known that 35% OF will suffer a second attack with in 5 years. it is decide to use a level of significance of 0.05 a total 225patients particepated by the taking daily asprien following their first heart attack. of those 45 had a second attack within 5 years

a, state the null and alternative haypothesis

b, what is the Z_crit

C,what is Z data

d, do you reject the null hypothesis?

e, state your conclusion using words in the pronlame

Calculate the following confidence intervals. The male confidence interval would be one calculation in the spreadsheet and the females would be a second calculation.  Mean for Females 7.666 and Standard Deviation 1.878. Mean for Males 7.764 and Standard Deviation 1.855. Total Males 17 and Total Females 18.  

1. Give and interpret the 95% confidence intervals for males and a second 95% confidence interval for females on the SLEEP variable. Which is wider and why?
2. Give and interpret the 99% confidence intervals for males and a second 99% confidence interval for females on the SLEEP variable. Which is wider and why?

All but the "face" cards (Kings, Queens, Jacks) have been removed from a regular deck of 52 playing cards

Draw two cards at random, without replacement. What is the probability that both cards are "Spades"? (Write answer as a fraction reduced to lowest terms) ___________

Based on your answer above, what are the Odds Against both cards being "Spades"? __________

Now draw a third card, again without replacement. What is the probability that this third card is a "Spade" GIVEN that NEITHER of the first two cards drawn are "Spades"? _____________

What would be your answer to the first question asked in this problem if the drawing had been done WITh replacement? ____________

Using diaries for many weeks, a study on the lifestyles of visually impaired students was conducted. The students kept track of many lifestyle variables including how many hours of sleep obtained on a typical day. Researchers found that visually impaired students averaged 9.63 hours of sleep, with a standard deviation of 2.92 hours. Assume that the number of hours of sleep for these visually impaired students is normally distributed.

(a) What is the probability that a visually impaired student gets at most 6.3 hours of sleep? Express your answer as a percent rounded to 2 decimal places. e.g. 1.23% Do not include the % symbol in your answer.

(b) What is the probability that a visually impaired student gets between 8 and 9.02 hours of sleep? Express your answer as a percent rounded to 2 decimal places. e.g. 1.23% Do not include the % symbol in your answer.

(c) What is the probability that a visually impaired student gets at least 8.2 hours of sleep? Express your answer as a percent rounded to 2 decimal places. e.g. 1.23% Do not include the % symbol in your answer.

(d) What is the sleep time that cuts off the top 33 % of sleep hours? Round your answer to 2 decimal places.

(e) If 400 visually impaired students were studied, how many students would you expect to have sleep times of more than 9.02 hours? Round to the nearest whole number.

(f) A school district wants to give additional assistance to visually impaired students with sleep times at the first quartile and lower. What would be the maximum sleep time to be recommended for additional assistance? Round your answer to 2 decimal places.

Assuming the probabilities of alternative prices for GM yellow corn are as stated in Exhibit 4,
calculate the expected change in profits from adopting our recommendation. (This is important since
it can be used to justify our consulting fees.)

Exhibit 4: Marketing and Price Analysis

Mercer Farms Group - Marketing Division

Background: The Marketing Division was asked to analyze the expected prices and probabilities for AA yellow corn and Genetically Modified (GM) yellow corn for the summer harvest.

Analysis: Estimating the future demand and supply of the commodity derives the projected market prices. The factors considered in the demand portion of this analysis include population growth, consumer preferences, and income. Relative prices of substitutes and complements were considered as static or unchanged. The supply portion of the analysis considered current input prices, existing technology, existing stocks on hand (domestic and foreign), and government policies (domestic and foreign). Exchange rate estimates were taken from our International Division’s current forecast.

Price Forecast:

AA Yellow Corn (domestic): Price per bushel: $ 5.00.

GM Yellow Corn (domestic):

Two alternative price scenarios should be considered. The demand acceptance of GM products in general is in question. There have been numerous reviews by governments all over the world, but particularly in Europe.

Scenario #1: Price of GM Yellow Corn (domestic): $ 5.50. Europe adopts few restrictions on the importation of GM products, but prohibits European production.

Scenario #2: Price of GM Yellow Corn (domestic): $ 4.70. Europe adopts heavy restrictions on the importation of GM products.

At this time, we consider the probabilities to be: Scenario #1: 60%; and Scenario #2: 40%.

The futures markets will have determined which price will occur before it is time to plant the summer crop.

he average daily dietary fiber intake was 16 grams per day. You believe that the average daily dietary fiber intake differs from 16 grams per day. What are the appropriate null and alternative hypotheses? Question 9 options: A) vs B) vs C) vs D) vs E) vs

Suppose the National Transportation Safety Board (NTSB) wants to examine the safety of compact cars, midsize cars, and full-size cars. It collects a sample of cars each of the cars types. The data below displays the frontal crash test performance percentages. Test whether there are statistical differences in the frontal crash test performance for each type of car.

What conclusions can we draw from the follow-up t-tests?  

There is/are a total of ["1", "2", "3", "4", "5", "6"] statistically significant difference(s) between car type pairs out of the follow-up t-tests.

Use the sample data and confidence level given below to complete parts (a) through (d).

A research institute poll asked respondents if they acted to annoy a bad driver. In the poll,

n equals 2528 comman=2528,

and

x equals 1199x=1199

who said that they honked. Use a

99 %99%

confidence level.

a) Find the best point estimate of the population proportion p.

b) Identify the value of the margin of error E

E=

c) Construct the confidence interval.

_<p<_

d) Write a statement that correctly interprets the confidence interval. Choose the correct answer below.

A.

One has

9999 %

confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.

B.

9999 %

of sample proportions will fall between the lower bound and the upper bound.

C.

There is a

9999 %

chance that the true value of the population proportion will fall between the lower bound and the upper bound.

D.

One has

9999 %

confidence that the sample proportion is equal to the population proportion.

Discuss the differences in a regression model between making the random error being multiplicative and making the random error being additive regarding how you approach estimation of the model coefficient(s), how you apply linearization for estimating the model coefficient(s), and how you obtain starting values for estimation of the model coefficient(s).

What are the pros and cons of using 98%confidence intervals?

9Do unregulated providers of child care in their homes follow different health and safety practices in different cities? A study looked at people who regularly provided care for someone else's children in poor areas of three cities. The numbers who required medical releases from parents to allow medical care in an emergency were 42 of 73 providers in Newark, New Jersey, 29 of 101 in Camden, New Jersey, and 48 of 107 in South Chicago, Illinois. A)Use the chi-square test to see if there are significant differences among the proportions of child care providers who require medical releases in the three cities. What do you conclude? B)How should the data be produced in order for your test to be valid? (In fact, the samples came in part from asking parents who were subjects in another study who provided their childcare. The author of the study, wisely did not use a statistical test. He wrote: Application of conventional statistical procedures appropriate for random samples may produce biased and misleading results.

Given two independent random samples with the following results: n1=11 x1=141 s1=21 n2=17 x2=116 s2=24 Use this data to find the 90% confidence interval for the true difference between the population means. Assume that the population variances are equal and that the two populations are normally distributed.

Step 1 of 3 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Step 2 of 3: Find the standard error of the sampling distribution to be used in construsting the confidence interval. round to the nearest whole number.

Step 3 of 3: construct the 99% confidence interval, round to the nearest whole number. (Lower endpoint, upper end point

A certain medical test is known to detect 38% of the people who are afflicted with the disease Y. If 10 people with the disease are administered the test, what is the probability that the test will show that:

All 10 have the disease, rounded to four decimal places?

At least 8 have the disease, rounded to four decimal places?

At most 4 have the disease, rounded to four decimal places?