In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities. What are the chances that a person who is murdered actually knew the murderer? The answer to this question explains why a lot of police detective work begins with relatives and friends of the victim! About 62% of people who are murdered actually knew the person who committed the murder.† Suppose that a detective file in New Orleans has 58 current unsolved murders. Find the following probabilities. (Round your answers to four decimal places.)

(a) at least 35 of the victims knew their murderers

(b) at most 48 of the victims knew their murderers

(c) fewer than 30 victims did not know their murderers

(d) more than 20 victims did not know their murderers

The table below summarizes data from a survey of a sample of women. Using a 0.05 significance​ level, and assuming that the sample sizes of 900 men and 400 women are​ predetermined, test the claim that the proportions of​ agree/disagree responses are the same for subjects interviewed by men and the subjects interviewed by women. Does it appear that the gender of the interviewer affected the responses of​ women?

   Man   Woman
Women who agree   605   322
Women who disagree   295   78

Identify the null and alternative hypotheses. Choose the correct answer below.

A.

Upper H 0H0​:

The proportions of​ agree/disagree responses are the same for the subjects interviewed by men and the subjects interviewed by women.

Upper H 1H1​:

The proportions are different.

B.

Upper H 0H0​:

The response of the subject and the gender of the subject are independent.

Upper H 1H1​:

The response of the subject and the gender of the subject are dependent.

C.

Upper H 0H0​:

The proportions of​ agree/disagree responses are different for the subjects interviewed by men and the subjects interviewed by women.

Upper H 1H1​:

The proportions are the same.

Compute the test statistic.

nothing

​(Round to three decimal places as​ needed.)

Find the critical​ value(s).

nothing

​(Round to three decimal places as needed. Use a comma to separate answers as​ needed.)

What is the conclusion based on the hypothesis​ test?

▼

Reject

Fail to reject

  

Upper H 0H0.

There

▼

is

is not

sufficient evidence to warrant rejection of the claim that the proportions of​ agree/disagree responses are the same for subjects interviewed by men and the subjects interviewed by women. It

▼

does not appear

appears

that the gender of the interviewer affected the responses of women.

Click to select your answer(s).

A woman sued a computer keyboard manufacturer, charging that her repetitive stress injuries were caused by the keyboard. The jury awarded about $3.5 million for pain and suffering, but the court then set aside that award as being unreasonable compensation. In making this determination, the court identified a "normative" group of 27 similar cases and specified a reasonable award as one within 2 standard deviations of the mean of the awards in the 27 cases. The 27 award amounts (in thousands of dollars) are in the table below.

33	60	75	115	135	140	149	150
232	290	340	410	600	750	750	750
1,050	1,100	1,139	1,150	1,200	1,200	1,250	1,574
1,700	1,825	2,000

What is the maximum possible amount (in thousands of dollars) that could be awarded under the "2-standard deviations rule"? (Round your answer to three decimal places.)

Below are 36 sorted ages of an acting award winner. Find Upper P 10 using the method presented in the textbook. 16,16,22,22,23,23,25,26,26,30,30,32,33,38,39,40,42,42,43,45,45,48,50,57,58,59,62,62,62,64,68,74,76,76,76,79

A​ case-control (or​ retrospective) study was conducted to investigate a relationship between the colors of helmets worn by motorcycle drivers and whether they are injured or killed in a crash. Results are given in the accompanying table. Using a 0.05 significance​ level, test the claim that injuries are independent of helmet color. Black White Yellow Red Blue Controls (not injured) 503 373 35 160 56 Cases (injured or killed) 215 122 10 66 28 Identify the null and alternative hypotheses. Choose the correct answer below. A. Upper H 0H0​: Whether a crash occurs and helmet color are independent Upper H 1H1​: Whether a crash occurs and helmet color are dependent B. Upper H 0H0​: Injuries and helmet color are independent Upper H 1H1​: Injuries and helmet color are dependent C. Upper H 0H0​: Whether a crash occurs and helmet color are dependent Upper H 1H1​: Whether a crash occurs and helmet color are independent D. Upper H 0H0​: Injuries and helmet color are dependent Upper H 1H1​: Injuries and helmet color are independent Compute the test statistic. nothing ​(Round to three decimal places as​ needed.) Find the critical​ value(s). nothing ​(Round to three decimal places as needed. Use a comma to separate answers as​ needed.) What is the conclusion based on the hypothesis​ test? ▼ Reject Fail to reject Upper H 0H0. There ▼ is not is sufficient evidence to warrant rejection of the claim that injuries are independent of helmet color. Click to select your answer(s).

The Lawnpoke Golf Association (LGA) has established rules that manufactures of golf equipment must meet for their products to be acceptable for LGA events. BatOutaHell Balls uses proprietary process to produce balls with mean distances of 295 yards. BatOutaHell is concerned that is the mean distance falls below 295 yards, the word will get out and sales will sage. Further, if the mean distance exceeds 295 yards, their balls may be rejected by LGA. Measurements of the distances are recorded in DATA. At ∝ =0.05, test the no action hypothesis that the balls have a mean distance of 295 yards.

Yards 293 275 280 290 273 306 287 301 309 285 274 282 294 283 304 297 294 290 283

A.The test statistic is 3.003 and the critical value is 1.734, therefore the test statistic is greater than the critical value of 1.734 and the null hypothesis is rejected. The distance is not 295 yards

B. The test statistic is 1.297 and the critical value is 1.734, therefore the test statistic is less than the critical value of 1.734 and the null hypothesis is not rejected. The distance is about 295 yards

C. The test statistics is 2.238 is greater than the critical value is 1.734, therefore H0 is rejected. It is reasonable to assume that the distance is not 295 yards.

D. The test statistic is 1.908 is greater than the critical value is 1.734, therefore H0 is rejected. It is reasonable to assume that the distance is not 295 yards.

Ohio Logistics manages the logistical activities for firms by matching companies that need products shipped with carriers that can provide the best rates and best service for the companies. Ohio Logistics is very concerned that its carriers deliver their customers' material on time, so it carefully monitors the percentage of on-time deliveries. The following table contains a list of the carriers used by Ohio Logistics and the corresponding on-time percentages for the current and previous year.

SEE BELOW to reference the data.

Carrier	Previous Year On-Time Deliveries (%)	Current Year On-Time Deliveries (%)
Blue Box Shipping	88.4	94.8
Cheetah LLC	89.3	91.8
Granite State Carriers	81.8	87.6
Honsin Limited	74.2	80.1
Jones Brothers	68.9	82.8
Minuteman Company	91.0	84.2
Rapid Response	78.8	70.9
Smith Logistics	84.3	88.7
Super Freight	92.1	86.8

(a)	Sort the carriers in descending order by their current year's percentage of on-time deliveries and answer the following questions.
	Which carrier is providing the best service in the current year?
	- Select your answer -Blue Box ShippingCheetah LLCGranite State CarriersHonsin LimitedJones BrothersMinuteman CompanyRapid ResponseSmith LogisticsSuper FreightItem 1
	Which carrier is providing the worst service in the current year?
	- Select your answer -Blue Box ShippingCheetah LLCGranite State CarriersHonsin LimitedJones BrothersMinuteman CompanyRapid ResponseSmith LogisticsSuper FreightItem 2
(b)	Calculate the change in percentage of on-time deliveries from the previous to the current year for each carrier. Use Excel's conditional formatting tool to choose the correct list of carrier whose on-time percentage decreased from the previous year to the current year.
	(i) Carrier Jones Brothers Minuteman Company Smith Logistics (ii) Carrier Rapid Response Minuteman Company Super Freight (iii) Carrier Honsin Limited Granite State Carriers Super Freight Rapid Response (iv) Carrier Minuteman Company Granite State Carriers Super Freight Cheetah LLC
	- Select your answer -List (i)List (ii)List (iii)List (iv)Item 3
(c)	Use Excel's conditional formatting tool to create data bars for the change in percentage of on-time deliveries from the previous year to the current year for each carrier calculated in part b. Choose the correct list of top 3 carriers whose on-time percentage increased from the previous year to the current year.
	(i) Carrier Jones Brothers Minuteman Company Smith Logistics (ii) Carrier Rapid Response Minuteman Company Super Freight (iii) Carrier Honsin Limited Granite State Carriers Super Freight (iv) Carrier Jones Brothers Blue Box Shipping Honsin Limited
	- Select your answer -List (i)List (ii)List (iii)List (iv)Item 4
(d)	Which carriers should Ohio Logistics try to use in the future? Why?
	The input in the box below will not be graded, but may be reviewed and considered by your instructor.
	DATA: Carrier Previous Year On-time Percentage Current Year On-time Percentage Blue Box Shipping 88.4% 94.8% Cheetah LLC 89.3% 91.8% Granite State Carriers 81.8% 87.6% Honsin Limited 74.2% 80.1% Jones Brothers 68.9% 82.8% Minuteman Company 91.0% 84.2% Rapid Response 78.8% 70.9% Smith Logistics 84.3% 88.7% Super Freight 92.1% 86.8%

Consider the following sample of observations on coating thickness for low-viscosity paint. 0.87 0.88 0.88 1.05 1.09 1.14 1.29 1.31 1.46 1.49 1.59 1.62 1.65 1.71 1.76 1.83 Assume that the distribution of coating thickness is normal (a normal probability plot strongly supports this assumption).

(c) Calculate a point estimate of the value that separates the largest 10% of all values in the thickness distribution from the remaining 90%. [Hint: Express what you are trying to estimate in terms of μ and σ.] (Round your answer to four decimal places.)

A soil scientist collected random soil samples from a field and measured the dissolved organic carbon (mg/L) level of each sample. The measurements are recorded in DissolvedOrganicCarbon.xlsx (Links to an external site.)Links to an external site.. The researcher would like to test if the mean dissolved organic carbon level in the field is less than 14 mg/L. Use a significance level of 0.05.

Part 1: Which hypothesis test is appropriate for this situation?  

The difference of means, using independent samples.   

The mean of differences, using dependent samples.   

One mean sigma known

What is the test statistic for the test mentioned above? (Show step by step)

What is the p-value for your test above? (Show step by step)

What is your conclusion based on the p-value above?

What is the 20th percentile of carbon level?

Dissolved Organic Carbon (mg/L)
14.367
11.958
12.038
12.875
14.690
10.297
9.901
11.325
12.105
16.519
6.185
24.275
20.792
15.197
9.879
18.820
13.296
9.845
10.906
8.587

R Code

2. Suppose X is a random variable with density f(x; ψ) = 2ψ-2 xI(0 ≤ x ≤ ψ), where ψ > 0 is a parameter.

(a) Draw a graph of the density when ψ = 3 in R.

(b) Write a function called genTri that generates n independent realizations of X. The function should take the arguments n and psi and return a vector of n realizations. Make a histogram of n = 104 realizations and compare the histogram to the density plot in the previous question.

(c) Derive an expression for E(X) – it will be a function of ψ. Confirm your expression when ψ = 3 using a Monte Carlo simulation.

(d) A 95% confidence interval for E(X) is given by Xbar ± t0.975,n−1 * S/√ n, where Xbar is the sample mean of n independent copies of X and t0.975,n−1 denotes the 97.5th percentile of the t-distribution with n − 1 degrees of freedom. For n = 5, 10, 15, 100 and ψ = 3, estimate the probability P(Xbar − t0.975,n−1 * S/√ n ≤ ψ ≤ Xbar + t0.975,n−1 * S/√ n) using a Monte Carlo simulation with 104 replications. Note: you can use the qt function in R to generate the respective t percentiles.

The following table shows the Myers-Briggs personality preference and area of study for a random sample of 519 college students. In the table, IN refers to introvert, intuitive; EN refers to extrovert, intuitive; IS refers to introvert, sensing; and ES refers to extrovert, sensing.

Myers-Briggs Preference	Arts & Science	Business	Allied Health	Row Total
IN	64	17	15	96
EN	84	44	26	154
IS	53	30	32	115
ES	70	44	40	154
Column Total	271	135	113	519

Use a chi-square test to determine if Myers-Briggs preference type is independent of area of study at the 0.05 level of significance.

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: Myers-Briggs type and area of study are not independent.
H₁: Myers-Briggs type and area of study are not independent.H₀: Myers-Briggs type and area of study are not independent.
H₁: Myers-Briggs type and area of study are independent.    H₀: Myers-Briggs type and area of study are independent.
H₁: Myers-Briggs type and area of study are not independent.H₀: Myers-Briggs type and area of study are independent.
H₁: Myers-Briggs type and area of study are independent.

(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?

YesNo    

What sampling distribution will you use?

Student's tbinomial    uniformchi-squarenormal

What are the degrees of freedom?


(c) Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.)

p-value > 0.1000.050 < p-value < 0.100    0.025 < p-value < 0.0500.010 < p-value < 0.0250.005 < p-value < 0.010p-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?

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, there is sufficient evidence to conclude that Myers-Briggs type and area of study are not independent.At the 5% level of significance, there is insufficient evidence to conclude that Myers-Briggs type and area of study are not independent.  

The file Sedans contains the overall miles per gallon (MPG) of 2013 midsized sedans:

38,26,30,26,25,27,22,27,39,24,24,26,25,23,25,26,31,26,37,22,29,25,33,21,21

Source: Data extracted from "ratings," Consumer Reports, April 2013, pp. 30-31.

A) Construct a 95% confidence interval estimate for the population mean MPG of 2013 family sedans, assuming a normal distribution.

B) Interpret the interval constructed in (a).

C) Compare the results in (a) to those in Problem 8.20(a).

How can I use confidence levels in statistics to solve a nurse patient ratio problem?

what informaitn would I need to do this? How would I collect data?

How would my information help me solve the problem?

From the list given to you for each of the following hypothesis testing situations, indicate the test you would use and explain the reason  

1. Pearsons Product-Moment Correlation Coefficient (Test #13)  

2. Spearman Rank Correlation (Test# 14)

3. Simple Linear Regression (Test# 15)  

4. ANOVA (Test # 16)

5. Kruskal-Wallis (Test # 17)

----

1. A researcher wants to know what the relationship is between how fifteen people do in a debate tournament (first place to fifteenth place), and how they do in a drama meet (also ordered from first to last)

2. You want to predict how many push-ups students will be able to do after participating in a fitness program from the number of push-ups they could do before the program starts.

3. You want to compare the mean numbers of push-ups done by 40 children who have attended one of three different fitness camps in order to see which fitness program is better.

4. A researcher wants to test the effects of 4 treatments on Poplar tree weights. The claim at the 5% significance level is that all four of the treatments produce the same median weight for poplar trees grown from seedlings. The hypotheses are:

Ho: M1 = M2 = M3 = M4

Ha: At least two medians differ from each other

Consider the following data regarding students' college GPAs and high school GPAs. The estimated regression equation is

Estimated College GPA=0.67+0.6551(High School GPA).Estimated College GPA=0.67+0.6551(High School GPA).

Compute the sum of squared errors (SSESSE) for the model. Round your answer to four decimal places.

GPAs

College GPA	High School GPA
2.022.02	3.293.29
2.812.81	3.113.11
2.532.53	3.303.30
3.763.76	4.974.97
3.083.08	3.003.00
3.963.96	3.873.87