(15 points) If in the previous example, the researchers also determine the weight class (medium and large) for each egg. 3 out 6 eggs in each housing system are medium and the other 3 are large.

Source DF SS MS F-Stat

HOUSING 0.8088

WTCLASS 0.06

Interaction 0.3078

Error N.A.

Total 4.2133   N.A. N.A.

1. Fill in the ANOVA table (total 13 missing values)

2. Are there main effects here? Do hypothesis testing at alpha=0.05. Does your conclusion

   consistent with question 2?

3. Is there evidence of interaction in the model? Do hypothesis testing at alpha=0.05. If

   there is interaction, give an example of possible interaction. If no interaction, explain what “no interaction” means

Epidemiologic case-control studies often report increased risk of an event given exposure, but we know that we can only calculate the odds ratio in a case-control study as opposed to a risk ratio.

Is it important to distinguish between a risk ratio and an odds ratio?

When does Odds Ratio approximate the Risk Ratio?

When does it approximate the Rate Ratio?

Please show steps in the calculation. Please make sure you include checking conditions for using the CLT.

The Human Resources (HR) Department of a certain college has asked all employees who were employed in 2018 to fill out a survey in December 2018. Three items on the survey were: “Your dental expense in 2018”, “Are you in a family with at least three other family members?”, and “Your medical expenses in 2018”. The manager of HR has randomly selected a sample of 169 surveys. Found that the sample average dental expense is $1600 per person with the sample standard deviation is $500. 70 of them are in a family of at least three other members. Also, the sample average medical expense is $2,450 per person and the sample standard deviation is $700.

1. The statistical inference will be made in parts (b), (c) and (d). Please clearly state the three parameters of interest.
2. Provide a 99% interval estimate of the average dental expenses per employee who was employed in the year 2018. please interpret your answer.

A research group conducted an extensive survey of 3180 wage and salaried workers on issues ranging from relationships with their bosses to household chores. The data were gathered through hour-long telephone interviews with a nationally representative sample. In response to the question, "What does success mean to you?" 1614 responded, "Personal satisfaction from doing a good job." Let p be the population proportion of all wage and salaried workers who would respond the same way to the stated question. Find a 90% confidence interval for p. (Round your answers to three decimal places.)

A classroom contains 10 students. A student committee of 3 (president, vice-president, and treasurer) must be selected. Harry Potter will serve only as president only if either of his friends Ron Weasley or Hermione Granger serve as vice president; otherwise he leaves the group. The others (including Ron and Hermione) have no such restrictions. What is the probability that Ron or Hermione become the president of this group?

Stat212 Statistical II B Spring 2020 Midterm 2 Name:

SHOW YOUR WORK TO EARN FULL CREDIT

1. (20 points) Susan Sound predicts that students will learn most effectively with a constant background sound, as opposed to an unpredictable sound. She randomly divides 15 students into 2 groups of 7 and 8 each. All students study a passage of text for 30 minutes. Those in group 1 study with background sound at a constant volume in the background. Those in group 2 study with noise that changes volume periodically. After studying, all students take a 10 point multiple choice test over the material. Their scores have the following test statistics.

Group Constant sound Random sound

Sample size Mean Variance

8 6 4.86
7 4 2.86

1. Construct the F statistic ANOVA table using data. Stat the hypotheses for ANIVA and use significance level 0.05 to test your hypotheses. What is the p-value of the test?

2. Explain whether you use can conclusion above to support Susan’s prediction or not? Explain why and how.

Find the t-critical values (tcrit) that form the rejection region for a two-tailed test with α = .05 for each of the following sample sizes.

a. N = 10 ________2.262____________

b. N = 17 _________2.120___________

c. N = 27 _________2.056____________

Problem 2) Willy Wonka’s Chocolate Factory produces four types of chocolate bars: • 30 percent of the chocolate bars are milk chocolate of which 10 percent have a golden ticket, • 15 percent of the chocolate bars are bitter chocolate of which 20 percent have a golden ticket, • 35 percent of the chocolate bars are mint chocolate of which 40 percent have a golden ticket, • 20 percent of the chocolate bars are vegemite chocolate of which 5 percent have a golden ticket, (1 point) (i) What is the probability that a randomly selected chocolate bar is milk chocolate? (2 points) (ii) What is the probability that a randomly selected chocolate bar has a golden ticket? (2 points) (iii) If a randomly selected chocolate bar has a golden ticket, what is the probability that it is bitter chocolate?

You wish to test the following claim (HaHa) at a significance level of α=0.02α=0.02.
Ho:μ1=μ2Ho:μ1=μ2
  Ha:μ1≠μ2Ha:μ1≠μ2
You believe both populations are normally distributed, but you do not know the standard deviations for either. However, you also have no reason to believe the variances of the two populations are not equal. You obtain the following two samples of data.

Sample #1      

sample2

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

What is the p-value for this sample? For this calculation, use the degrees of freedom reported from the technology you are using. (Report answer accurate to four decimal places.)
p-value =

The p-value is...

• less than (or equal to) αα
• greater than αα

This test statistic leads to a decision to...

• reject the null
• accept the null
• fail to reject the null

As such, the final conclusion is that...

• There is sufficient evidence to warrant rejection of the claim that the first population mean is not equal to the second population mean.
• There is not sufficient evidence to warrant rejection of the claim that the first population mean is not equal to the second population mean.
• The sample data support the claim that the first population mean is not equal to the second population mean.
• There is not sufficient sample evidence to support the claim that the first population mean is not equal to the second population mean.

Exercise 3.8.18: How many ways are there to choose five cards from 52 so that you have all the aces? How many ways are there to choose five cards from 52 so that you have four cards with the same number?

Exercise 3.8.20: How many ways are there to choose five cards from 52 so that you have two cards with the same number? Hint: This problem is more tricky than you think.

According to a marketing​ website, adults in a certain country average 5454 minutes per day on mobile devices this year. Assume that minutes per day on mobile devices follow the normal distribution and has a standard deviation of 1212 minutes. Complete parts a through d below.

A. What is the probability that the amount of time spent today on mobile devices by an adult is less than 63 ​minutes?

(Round to four decimal places as​ needed.)

B. What is the probability that the amount of time spent today on mobile devices by an adult is more than 42 ​minutes?

(Round to four decimal places as​ needed.)

C. What is the probability that the amount of time spent today on mobile devices by an adult is between 32 and 49 minutes?

(Round to four decimal places as​ needed.)

D. What amount of time spent today on mobile devices by an adult represents the 85th ​percentile?

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

A random sample of 90 observations produced a mean of 32.4 from a population with a normal distribution and a standard deviation ?=2.98.

(a) Find a 90% confidence interval for μ
≤?≤

(b) Find a 95% confidence interval for μ
≤?≤

(c) Find a 99% confidence interval for μ
≤?≤

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.

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 different.    H₀: The distributions are the same.
H₁: The distributions are the same.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?

YesNo    

What sampling distribution will you use?

chi-squareStudent's t    normalbinomialuniform

What are the degrees of freedom?


(c) Estimate the P-value of the sample test statistic.

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 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.    

Suppose a distribution has a mean of 100 and a standard deviation of 20. Further suppose that random samples of size n = 100 are taken with replacement from this distribution. The mean of the sampling distribution of sample means is mu subscript x with bar on top end subscript = and the standard deviation of the sampling distribution of sample means is sigma subscript top enclose x end subscript = .

A law firm specializing in corporate law has 4 lawyers. Because of various reasons, on any given day, the probabilities that each lawyer will be attending a trial is:  0.90, 0.90, 0.85, 0.80. Note that a lawyer attending a trial is independent of any other lawyer attending a trial

a.  Find the probability distribution for the number of lawyers who has a trial on a given day.

b. Find the expected number of lawyers attending to a trial on a given day.

c. Find the standard deviation of the number of lawyers attending to a trial on a given day. Round your answer to four decimal points.