One possible explanation for why some birds migrate and others maintain year-round residency in a single location is intelligence. Specifically, birds with small brains, relative to their body size, are simply not smart enough to find food during the winter and must migrate to warmer climates where food is easily available (Sol, Lefebvre, & Rodriguez-Teijeiro, 2005). Birds with bigger brains, on the other hand, are more creative and can find food even when the weather turns harsh. Following are hypothetical data similar to the actual research results. The numbers represent relative brain size for the individual birds in each sample.

Step by step process on SPSS for One-Way ANOVA

a. Use an analysis of variance with alpha .05 to determine whether there are any significant differences among the three groups of birds.

b. Compute n2, the percentage of variance explained by the differences, between groups, for these data.

A psychologist would like to examine how the rate of presentation affects people’s ability to memorize a list of words. A list of 20 words is prepared. For one group of participants the list is presented at the rate of one word every ½ second. The next group gets one word every second. The third group has one word every 2 seconds, and the fourth group has one word every 3 seconds. After the list is presented, the psychologist asks each person to recall the entire list. The dependent variable is the number of errors in recall. The data from this experiment are as follows:

Step by step on SPSS

a. Can the psychologist conclude that the rate of presentation has a significant effect on memory? Test at the .05 level.

b. Use the Tukey HSD test to determine which rates of presentation are statistically different and which are not.

A radio tube inserted into a system has probability 0.2 of lasting 500 hours. 20 tubes are tested.

1. Find probability that exactly that 'k' of these tubes will last more than 500 hours.

2. Find probability that number of tubes that last more than 500 hours will fall between 12 and 17.

3. Sketch the CDF of the random variable that describes the random phenomenon

Also: a die is rolled 120 times. Find probability that 35 or more sixes will be rolled.

Show all steps, thank you.

Abstract 2. Mann JR and McDermott S. Are maternal genitourinary infection and pre-eclampsia associated with ADHD in school-aged children? J Atten Disord 2011;15(8):667-73.

OBJECTIVE: To investigate the hypothesis that maternal genitourinary infection (GU) infection is associated with increased risk of ADHD. METHOD: The authors obtained linked Medicaid billing data for pregnant women and their children in South Carolina, with births from 1996 through 2002 and follow-up data through 2008. Maternal GU infections and pre-eclampsia were identified on the basis of diagnoses made during pregnancy, and cases of ADHD were identified on the basis of diagnoses made in the child's Medicaid file. RESULTS: There were 84,721 children in the data set used for analyses. Maternal genitourinary infection was associated with significantly increased odds of ADHD (OR = 1.29, 95% CI = 1.23-1.35). Pre-eclampsia was also associated with increased risk (OR = 1.19, 95% CI = 1.07-1.32). Children whose mothers had both GU infection and pre-eclampsia were 53% more likely to have ADHD, compared to those with neither exposure. When we examined specific infection diagnoses, chlamydia/nongonococcal urethritis, trichomoniasis, urinary tract infection, and candidiasis were associated with increased risk of ADHD, whereas gonorrhea was not. DISCUSSION: Maternal GU infection appeared to be associated with increased risk of ADHD, and based on the findings it was concluded that further research is needed to describe the mechanism(s) underlying the association.

The authors report “pre-eclampsia was also associated with increased risk (OR=1.19, 95% CI=1.07-1.32” of ADHD. Based only on this statement, which of the following is most likely NOT responsible for this observed result?

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

a) The test statistic of 2.238 is greater than the critical value is 2.101 therefore HO is rejected. it is reasonable to assume that the distance is not 295 yards.

b) The test statistic of 3.003 and the critical value is 2.101, therefore the test statistic is greater than the critical value of 2.101 and the null hypothesis is rejected. The distance is not 295 yards

c) The test statistic of 1.908 is less than the critical value of 2.101 therefore HO is not in the area of rejection. It is reasonto assume that the distance is about 295 yards.

d) The test statistic is 1.297 and the critical value is 2.101 therefore the test statistic is less than the critical value and the null hypothesis is not rejected. The distance is about 295 yards.

Let x represent the dollar amount spent on supermarket impulse buying in a 10-minute (unplanned) shopping interval. Based on a certain article, the mean of the xdistribution is about $26 and the estimated standard deviation is about $7.

(a) Consider a random sample of n = 60 customers, each of whom has 10 minutes of unplanned shopping time in a supermarket. From the central limit theorem, what can you say about the probability distribution of x, the average amount spent by these customers due to impulse buying? What are the mean and standard deviation of the x distribution?

-The sampling distribution of x is approximately normal with mean μ_x = 26 and standard error σ_x = $7.

The sampling distribution of x is not normal.     

-The sampling distribution of x is approximately normal with mean μ_x = 26 and standard error σ_x = $0.90.

-The sampling distribution of x is approximately normal with mean μ_x = 26 and standard error σ_x = $0.12.

Is it necessary to make any assumption about the x distribution? Explain your answer.

-It is not necessary to make any assumption about the x distribution because n is large.I

-t is necessary to assume that x has a large distribution.     

-It is not necessary to make any assumption about the x distribution because μ is large.

-It is necessary to assume that x has an approximately normal distribution.

(b) What is the probability that x is between $24 and $28? (Round your answer to four decimal places.)


(c) Let us assume that x has a distribution that is approximately normal. What is the probability that x is between $24 and $28? (Round your answer to four decimal places.)


(d) In part (b), we used x, the average amount spent, computed for 60 customers. In part (c), we used x, the amount spent by only one customer. The answers to parts (b) and (c) are very different. Why would this happen?

-The standard deviation is larger for the x distribution than it is for the x distribution.

-The standard deviation is smaller for the x distribution than it is for the x distribution.     

-The mean is larger for the x distribution than it is for the x distribution.

-The x distribution is approximately normal while the x distribution is not normal.

-The sample size is smaller for the x distribution than it is for the x distribution.

In this example, x is a much more predictable or reliable statistic than x. Consider that almost all marketing strategies and sales pitches are designed for the average customer and not the individual customer. How does the central limit theorem tell us that the average customer is much more predictable than the individual customer?

-The central limit theorem tells us that the standard deviation of the sample mean is much smaller than the population standard deviation. Thus, the average customer is more predictable than the individual customer.

-The central limit theorem tells us that small sample sizes have small standard deviations on average. Thus, the average customer is more predictable than the individual customer.    

In Born together—Reared apart: the Landmark Minnesota twin study (2012), Nancy Segal discusses the efforts of research psychologists at the University of Minnesota to understand similarities and differences between twins by studying sets of twins who were raised separately. The Excel Online file below contains critical reading SAT scores for several pairs of identical twins (twins who share all of their genes), one of whom was raised in a family with no other children (no siblings) and one of whom was raised in a family with other children (with siblings). Construct a spreadsheet to answer the following questions.

Open spreadsheet

a. What is the mean difference between the critical reading SAT scores for the twins raised with no siblings and the twins raised with siblings?

(to 2 decimals)

b. Provide a 90% confidence interval estimate of the mean difference between the critical reading SAT scores for the twins raised with no siblings and the twins raised with siblings.

(, ) (to 2 decimals)

c. Conduct a hypothesis test of equality of the critical reading SAT scores for the twins raised with no siblings and the twins raised with siblings.

-value is  (to 4 decimals)

At , what is your conclusion?

_________Can concludeCannot conclude that there is a difference between the mean scores for the no sibling and with sibling groups.

Cincinnati Paint Company sells quality brands of paints through hardware stores throughout the United States. The company maintains a large sales force who call on existing customers and look for new business. The national sales manager is investigating the relationship between the number of sales calls made and the miles driven by the sales representative. Also, do the sales representatives who drive the most miles and make the most calls necessarily earn the most in sales commissions? To investigate, the vice president of sales selected a sample of 25 sales representatives and determined:

• The amount earned in commissions last month (y)
• The number of miles driven last month (x₁)
• The number of sales calls made last month (x₂)

Develop a regression equation including an interaction term. (Negative amount should be indicated by a minus sign. Round your answers to 3 decimal places.)

Complete the following table. (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.)

Compute the value of the test statistic corresponding to the interaction term. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)

Value of the test statistic

At the 0.05 significance level is there a significant interaction between the number of sales calls and the miles driven?

Given that xx is a random variable having a Poisson distribution, compute the following:

(a)  P(x=1)P(x=1) when μ=4.5μ=4.5
P(x)=P(x)=

(b)  P(x≤8)P(x≤8)when μ=0.5μ=0.5
P(x)=P(x)=

(c)  P(x>7)P(x>7) when μ=4μ=4
P(x)=P(x)=

(d)  P(x<1)P(x<1) when μ=1μ=1
P(x)=P(x)=

A group of college students want to have a party. They need to decide if they want to have it at the Beach (B) or at the Park (P) or in a Warehouse (W). They were ask to list in order their 1^st, 2^nd, and 3^rd choice of where they want to have the party. Use the following table and answer the following questions. SHOW YOUR WORK to receive full credit.

a. How many votes were cast?

b. Use the plurality method to determine the winner.

c. Use the instant runoff method to determine the winner.

d. Use the Borda count method to determine the winner.

We are interested in looking at ticket prices of MLB games. It is known from past information that the average price is $26.30, with a population standard deviation of $2.32. Suppose we take a sample of the last 5 years and find that the average ticket price is $29.94. We are interested in seeing if the average price of tickets has significantly increased. Use alpha=.10. Find the 90% confidence interval for ticket price.

We are interested in looking at ticket prices of MLB games. It is known from past information that the average price is $26.30, with a population standard deviation of $2.32. Suppose we take a sample of the last 5 years and find that the average ticket price is $29.94. We are interested in seeing if the average price of tickets has significantly increased. Use alpha=.10. The null and alternative hypotheses are:

We are interested in looking at ticket prices of MLB games. It is known from past information that the average price is $26.30, with a population standard deviation of $2.32. Suppose we take a sample of the last 5 years and find that the average ticket price is $29.94. We are interested in seeing if the average price of tickets has significantly increased. Use alpha=.10. What is the critical value?

We are interested in looking at ticket prices of MLB games. It is known from past information that the average price is $26.30, with a population standard deviation of $2.32. Suppose we take a sample of the last 5 years and find that the average ticket price is $29.94. We are interested in seeing if the average price of tickets has significantly increased. Use alpha=.10. What is the test statistic?

We are interested in looking at ticket prices of MLB games. It is known from past information that the average price is $26.30, with a population standard deviation of $2.32. Suppose we take a sample of the last 5 years and find that the average ticket price is $29.94. We are interested in seeing if the average price of tickets has significantly increased. Use alpha=.10. What is the p-value?

Altira Corporation uses a periodic inventory system. The following information related to its merchandise inventory during the month of August 2018 is available:

Determine the inventory balance Altira would report in its August 31, 2018, balance sheet and the cost of goods sold it would report in its August 2018 income statement using the FIFO method. (Round "Cost per Unit" to 2 decimal places.)

HINT: The boxes with the "!" are the ones that need to be filled in

It is known from past information that the probability of failing to finish an Ironman (called a DNF) is 15%. Suppose we take a sample of 100 Ironman races over the past few years and find that the DNF percentage is 12%. We are interested in seeing if there has been a significant decrease in the number of DNFs. Use alpha=0.05. Find the 95% confidence interval for the percentage of DNFs.

It is known from past information that the probability of failing to finish an Ironman (called a DNF) is 15%. Suppose we take a sample of 100 Ironman races over the past few years and find that the DNF percentage is 12%. We are interested in seeing if there has been a significant decrease in the number of DNFs. Use alpha=0.05. The null and the alternative hypotheses are:

It is known from past information that the probability of failing to finish an Ironman (called a DNF) is 15%. Suppose we take a sample of 100 Ironman races over the past few years and find that the DNF percentage is 12%. We are interested in seeing if there has been a significant decrease in the number of DNFs. Use alpha=0.05. The test statistic is:

It is known from past information that the probability of failing to finish an Ironman (called a DNF) is 15%. Suppose we take a sample of 100 Ironman races over the past few years and find that the DNF percentage is 12%. We are interested in seeing if there has been a significant decrease in the number of DNFs. Use alpha=0.05. The p-value is:

It is known from past information that the probability of failing to finish an Ironman (called a DNF) is 15%. Suppose we take a sample of 100 Ironman races over the past few years and find that the DNF percentage is 12%. We are interested in seeing if there has been a significant decrease in the number of DNFs. Use alpha=0.05. Assuming no prior estimate of p is available, what is the sample size that would need to be taken if we wanted to have a margin of error of 0.08 or less with 95% confidence?

An athletic footwear company is attempting to estimate the sales that will result from a television advertisement campaign of its new athletic shoe. The contribution to earnings from each pair of shoes sold is $40. Suppose that the probability that a television viewer will watch the advertisement (as opposed to turn his/her attention elsewhere) is 0.40. Furthermore, suppose that 1% of viewers who watch the advertisement on a local television channel will buy a pair of shoes. The company can buy television advertising time in one of the time slots according to Table below:
Television advertising costs and viewers

(a) Suppose that the company decides to buy one minute of advertising time. Which time slot would yield the highest expected contribution to earnings net of costs? What is the total expected contribution to earnings resulting from the advertisement?
(b) Suppose the company decides to buy two one-minute advertisements in different time slots. Which two different time slots should the company purchase to maximize the expected contribution to earnings? What is the total expected contribution to earnings resulting from these two advertisements?