A quality survey asked recent customers of their experience at a local department store. One question asked for the customers rating on their service using categorical responses of average, outstanding, and exceptional. Another question asked for the applicant’s education level with categorical responses of Some HS, HS Grad, Some College, and College Grad. The sample data below are for 700 customers who recently visited the department store. Education Quality Rating Some HS HS Grad Some College College Grad Average 55 80 50 35 Outstanding 60 105 65 70 Exceptional 35 65 35 45 Using a level of significance of 0.01, is there evidence to suggest that the customer’s Education level and Quality Rating are independent? In other words, is there a relationship or is there NO relationship between Education and Quality Rating? a. State the Null and Alternative hypothesis. b. What is the statistic you would use to analyze this? c. State your decision rule: d. Show your calculation: e. What is your conclusion? Quality Rating and Education level are independent OR Quality Rating and Education level are NOT independent

5. Four different paints are advertised to have the same drying times. To verify the manufacturer’s claim, seven samples were tested for each of the paints. The time in minutes until the paint was dry enough for a second coat to be applied was recorded. Below are the results which may be imported into MSExcel for analysis: (Assume the populations are normally distributed, the populations are independent and the population variances are equal) Paint 1 Paint 2 Paint 3 Paint 4 120 120 117 128 112 130 122 131 121 121 123 131 118 126 115 129 118 126 123 127 121 114 126 126 118 117 126 137 a. Write the Null and Alternative hypothesis to test whether there is a difference in dry time between the samples of each paint? b. What statistic would you use to analyze this? c. At a 0.01 level of significance, what would be your decision rule? d. From your analysis, is there a difference between drying times? e. If there was a difference in dry times how would you determine which paint has the different drying time? (i.e. what formula would you use and what is your decision criteria?) (this is not asking for a calculation)

In​ 2008, the per capita consumption of soft drinks in Country A was reported to be 19.35 gallons. Assume that the per capita consumption of soft drinks in Country A is approximately normally​ distributed, with a mean of 19.35 gallons and a standard deviation of 4 gallons. Complete parts​ (a) through​ (d) below.

a. What is the probability that someone in Country A consumed more than 14 gallons of soft drinks in​ 2008? The probability is nothing. ​(Round to four decimal places as​ needed.)

b. What is the probability that someone in Country A consumed between 9 and 10 gallons of soft drinks in​ 2008? The probability is nothing. ​(Round to four decimal places as​ needed.)

c. What is the probability that someone in Country A consumed less than 10 gallons of soft drinks in​ 2008? The probability is nothing. ​(Round to four decimal places as​ needed.)

d. 97​% of the people in Country A consumed less than how many gallons of soft​ drinks? The probability is 97​% that someone in Country A consumed less than nothing gallons of soft drinks. ​(Round to two decimal places as​ needed.)

Data from 1991 General Social Survey classify a sample of Americans according to their gender and their opinion about afterlife (example from A. Agresti, 1996, “Introduction to categorical data analysis”). The opinions about afterlife were classified into two categories: Yes and No (or undecided). For example, for the females in the sample - 435 said that they believed in an afterlife and 147 said that they did not or were undecided.

Estimate the proportion of females who believed in an afterlife (Use a 95% Confidence Interval).

Test hypothesis that the majority of females (that is, more than 50% females) believed in an afterlife.

- Using a z-score test

Using c² test

4. Are biomedical engineering salaries in Miami less than those in Minnesota? Salary data show that staff biomedical engineers in Miami earn less than those in Minnesota. Suppose that in a follow-up study of 35 staff engineers in Miami and 45 staff engineers in Minnesota you obtain the following results: Miami Minnesota n1 = 35 n2 = 45 1 = $64, 150 1 = $65, 450 s1 = $2000 s2 = 2500 a. Formulate a hypothesis statement so that if the null hypothesis is rejected, we can conclude that Miami biomedical engineering salaries are significantly lower than those in Minnesota. b. The standard deviations for both populations can be assumed equal, write out the statistic you would use. c. What would be your decision rule? (Use α = 0.05) d. What is the value of your statistic? e. What is your p-value? f. What is your conclusion?

(SHOW WORK)

A television sports commentator wants to estimate the proportion of citizens who​ "follow professional​ football." Complete parts​ (a) through​ (c).

​(a) What sample size should be obtained if he wants to be within

44

percentage points with

9494​%

confidence if he uses an estimate of

4848​%

obtained from a​ poll?The sample size is

nothing.

​(Round up to the nearest​ integer.)​(b) What sample size should be obtained if he wants to be within

44

percentage points with

9494​%

confidence if he does not use any prior​ estimates?The sample size is

nothing.

​(Round up to the nearest​ integer.)

​(c) Why are the results from parts​ (a) and​ (b) so​ close?

A.The results are close because the margin of error

44​%

is less than​ 5%.

B.The results are close because

0.48 left parenthesis 1 minus 0.48 right parenthesis equals0.48(1−0.48)=0.24960.2496

is very close to 0.25.

C.The results are close because the confidence

9494​%

is close to​ 100%.

You are preparing some sweet potato pie for your annual Thanksgiving feast. The store sells sweet potatoes in packages of 6. From prior experience you have found that the package will contain 0 spoiled sweet potatoes about 65% of the time, 1 spoiled sweet potato 25% of the time and 2 spoiled sweet potatoes the rest of the time. Conduct a simulation to estimate the number of packages of sweet potatoes you need to purchase to have three dozen (36) unspoiled sweet potatoes.

Part 1 of 3: (6 pts)

Describe in detail and in paragraph form how you will use the random numbers provided from a random number table (in part 2) to conduct 2 trials of this simulation. Be sure to include all of the first four steps of a simulation. Steps 5 and 6 are part 2 and step 7 are part 3 of this question.

1. Identify the component to be repeated.

2. Explain how you will model the outcome.

3. Explain in detail how you will simulate the trial.

4. State clearly what the response variable is.

Part 2 of 3: (3 pts)

Use the random number table to complete 2 trials. Analyze your response variable.

Trial#1:   41  23  19  98  75  08  63  29  10

Trial #2: 88 26 95  69  57  71  02 62 34

Part 3 of 3: (1 pt)

Give your conclusion based on your simulation results.

Each observation in a random sample of 106 bicycle accidents resulting in death was classified according to the day of the week on which the accident occurred. Data consistent with information are given in the following table. Based on these data, is it reasonable to conclude that the proportion of accidents is not the same for all days of the week? Use α = 0.05. (Round your answer to two decimal places.)

χ² =  

P-value interval

p < 0.001

0.001 ≤ p < 0.01    

0.01 ≤ p < 0.05

0.05 ≤ p < 0.10

p ≥ 0.10

The proportion of accidents is  ---Select--- the same, not the same for all days.

Could you please explain to me when I should use the two proportion tests?

A two proportion f test tests variance/standard deviation. So if it is testing how much the two proportions vary then isn't it doing what a linear regression t test does?

Whats the difference between a linear regression t test and goodness of fit test statistic?

Is there such thing as a two proportion chi squared test or is that just an f test?

The only difference between a two proportion t test and a two proportion z test is the knowing of a population standard deviation so that is good.

As you can see I'm all jumbled up.

Are very young infants more likely to imitate actions that are modeled by a person or simulated by an object? This question was the basis of a research study. One action examined was mouth opening. This action was modeled repeatedly by either a person or a doll, and the number of times that the infant imitated the behavior was recorded. Twenty-seven infants participated, with 12 exposed to a human model and 15 exposed to the doll. Summary values are shown below.

Is there sufficient evidence to conclude that the mean number of imitations is higher for infants who watch a human model than for infants who watch a doll? Test the relevant hypotheses using a 0.01 significance level. (Use a statistical computer package to calculate the P-value. Use μ_Person − μ_Doll. Round your test statistic to two decimal places, your df down to the nearest whole number, and your P-value to three decimal places.)

t=

df=

P-value=

State your conclusion.

We reject H₀. We do not have convincing evidence that the mean number of imitations is higher for infants who watch a human model than for infants who watch a doll.

We do not reject H₀. We have convincing evidence that the mean number of imitations is higher for infants who watch a human model than for infants who watch a doll.    

We reject H₀. We have convincing evidence that the mean number of imitations is higher for infants who watch a human model than for infants who watch a doll.We do not reject H₀.

We do not have convincing evidence that the mean number of imitations is higher for infants who watch a human model than for infants who watch a doll.

CNNBC recently reported that the mean annual cost of auto insurance is 988 dollars. Assume the standard deviation is 105 dollars. You will use a simple random sample of 110 auto insurance policies.

Find the probability that a single randomly selected policy has a mean value between 994 and 1008 dollars.
P(994 < X < 1008) =

Find the probability that a random sample of size n=110n=110 has a mean value between 994 and 1008 dollars.
P(994 < M < 1008) =

Many different manufacturers sell residential gas ranges. The cost in dollars of four gas ranges is given below.

529 664 709 800

1. suppose a random sample of size two is selected from this population without replacement. Find the sampling distribution of the sample mean.
2. Suppose a random sample of size is selected from this population with replacement. Find the sampling distribution of the sample mean
3. How are these two distributions similar? How are they different?

Based on historical data, your manager believes that 44% of the company's orders come from first-time customers. A random sample of 141 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.26 and 0.48?

Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations. Answer = (Enter your answer as a number accurate to 4 decimal places.)

In a study of 1910 schoolchildren in Australia, 1050 children indicated that they normally watch TV before school in the morning. (Interestingly, only 35% of the parents said their children watched TV before school!)

(a)

Construct a 95% confidence interval for the true proportion of Australian children who say they watch TV before school. (Round your answers to three decimal places.)

(_____,________)
What assumption about the sample must be true for the method used to construct the interval to be valid?(b)

The 1910 schoolchildren used in the study formed a random sample from the population of children in Australia who normally watch TV before school in the morning.

The 1050 children who indicated that they normally watch TV before school in the morning formed a random sample from the population of schoolchildren in Australia.    

The 1910 schoolchildren used in the study formed a random sample from the population of schoolchildren in Australia.

The 1050 children who indicated that they normally watch TV before school in the morning formed a random sample from the population of children in Australia who normally watch TV before school in the morning.

How much money do people spend on graduation gifts? In 2007, a federation surveyed 2415 consumers who reported that they bought one or more graduation gifts that year. The sample was selected in a way designed to produce a sample representative of adult Americans who purchased graduation gifts in 2007. For this sample, the mean amount spent per gift was $58.15. Suppose that the sample standard deviation was $20. Construct a 98% confidence interval for the mean amount of money spent per graduation gift in 2007. (Round your answers to three decimal places.)
(  ,  )

Interpret the interval.

We are 98% confident that the mean amount of graduation money spent was within this interval.

We are 98% confident that the mean amount of money spent per graduation gift in 2007 was within this interval.    

We are confident that the mean amount of money spent per graduation gift in 2007 was within this interval 98% of the time.

We are confident that 98% of the amount of money spent per graduation gift in 2007 was within this interval.