Questions
In a random sample of 88 ears of corn, farmer Carl finds that 9 of them...

In a random sample of 88 ears of corn, farmer Carl finds that 9 of them have worms. He wants to find the 99% confidence interval for the proportion of all his corn that has worms.

(a) What is the point estimate for the proportion of all of Carl's corn that has worms? Round your answer to 3 decimal places.

(b) What is the critical value of z (denoted zα/2) for a 99% confidence interval? Use the value from the table or, if using software, round to 2 decimal places. zα/2 =

(c) What is the margin of error (E) for a 99% confidence interval? Round your answer to 3 decimal places. E =

(d) Construct the 99% confidence interval for the proportion of all of Carl's corn that has worms. Round your answers to 3 decimal places. < p <

(e) Based on your answer to part (d), are you 99% confident that less than 22% of Carl's corn has worms?

No, because 0.22 is above the upper limit of the confidence interval.

Yes, because 0.22 is below the upper limit of the confidence interval.

Yes, because 0.22 is above the upper limit of the confidence interval.

No, because 0.22 is below the upper limit of the confidence interval. Additional Materials

In: Math

SOLVE WITH SPSS ONLY We wish to assess the effect of three different track surfaces on...

SOLVE WITH SPSS ONLY

We wish to assess the effect of three different track surfaces on sprinter speed. Six world-class sprinters are asked to run five 100m dashes on each of the three track surfaces. Their average times are recorded below. USE SPSS NO EXCEL NO HAND

Surface 1 Surface 2 Surface 3

Sprinter 1 9.85 10.00 10.04

Sprinter 2 9.90 10.07 10.16

Sprinter 39.89 9.99 10.17

Sprinter 4 9.88 9.98 10.04

Sprinter 5 9.81 10.03 10.10

Sprinter 6 9.81 9.95 10.12

Using the sprinters as blocks, discuss the differences between the track surfaces as suggested by SPSS. Give a statement to be tested, identify the random variables involved and the assumptions you make about them, state the hypotheses to be tested, ask SPSS to run the analysis for you, including a post hoc, and then discuss the outcome. Describe the critical region(s) upon which you base your decisions. Include any SPSS output in your discussion.

In: Math

Almost all medical schools in the United States require students to take the Medical College Admission...

Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate the mean score μ of those who took the MCAT on your campus, you will obtain the scores of an SRS of students. The scores follow a Normal distribution, and from published information you know that the standard deviation is 10.4 . Suppose that, unknown to you, the mean score of those taking the MCAT on your campus is 495 .

In answering the questions, use z‑scores rounded to two decimal places.

(a) If you choose one student at random, what is the probability that the student's score is between 490 and 500 ? Use Table A, or software to calculate your answer.

(Enter your answer rounded to four decimal places.)

probability:

(b) You sample 25 students. What is the standard deviation of the sampling distribution of their average score ?¯? (Enter your answer rounded to two decimal places.)

standard deviation:

(c) What is the probability that the mean score of your sample is between 490 and 500 ? (Enter your answer rounded to four decimal places.)

probability:

STATE: How heavy a load (in pounds) is needed to pull apart pieces of Douglas fir 44 inches long and 1.51.5 inches square? Given are data from students doing a laboratory exercise.

33,190 31,860 32,590 26,520 33,280
32,320 33,020 32,030 30,460 32,700
23,040 30,9303 32,720 33,650 32,340
24,050 30,170 31,300 28,730 31,920

To access the complete data set, click the link for your preferred software format:

Excel  Minitab  JMP  SPSS TI  R  Mac-TXT   PC-TXT  CSV CrunchIt!

We are willing to regard the wood pieces prepared for the lab session as an SRS of all similar pieces of Douglas fir. Engineers also commonly assume that characteristics of materials vary Normally. Suppose that the strength of pieces of wood like these follows a Normal distribution with standard deviation 3000 pounds.

PLAN: We will estimate μ by giving a 90% confidence interval.

SOLVE: Find the sample mean ?¯ . (Enter your answer rounded to the nearest whole number.)

?¯=

Give a 90% confidence interval, [low,high] , for the mean load required to pull the wood apart. (Enter your answers rounded to the nearest whole number.)

???=

ℎ??ℎ=

In: Math

Part I: Between-Groups Design In the between-groups design, researchers were interested in whether cholesterol levels would...

Part I: Between-Groups Design

In the between-groups design, researchers were interested in whether cholesterol levels would differ depending on diet. Twenty participants were randomly assigned to one of two different groups. Group A was assigned a diet rich in fruits and vegetables and with no trans fats. Group B participants were asked to follow their normal diets, which contained varying levels of trans fats depending on the individual. After one month, blood samples were drawn and the following levels of cholesterol were obtained:

Participant

Blood Cholesterol

Diet

1

129

Healthy

2

98

Healthy

3

150

Healthy

4

75

Healthy

5

135

Healthy

6

175

Healthy

7

115

Healthy

8

103

Healthy

9

156

Healthy

10

143

Healthy

11

239

Normal

12

500

Normal

13

350

Normal

14

468

Normal

15

198

Normal

16

213

Normal

17

225

Normal

18

175

Normal

19

560

Normal

20

289

Normal

In 2 to 3 sentences in a Microsoft Word document, answer the following questions:

  • What is the independent variable in this study?
  • What are the levels of that independent variable?
  • What is the dependent variable?

Part II: Within-Subjects Design

In the within-subjects design, researchers were interested in whether participants could lower their cholesterol levels by changing from a diet higher in trans fats to one with no trans fats. Ten research participants were selected. A baseline measure of cholesterol was taken from each. They were then put on a diet rich in fruits and vegetables and devoid of trans fats for one month. At the end of that month, blood cholesterol was again measured and the following results were obtained:

Participant

Blood Cholesterol

Diet

1

129

Baseline

1

98

Healthy

2

150

Baseline

2

75

Healthy

3

175

Baseline

3

135

Healthy

4

115

Baseline

4

103

Healthy

5

156

Baseline

5

143

Healthy

6

500

Baseline

6

450

Healthy

7

468

Baseline

7

350

Healthy

8

198

Baseline

8

213

Healthy

9

225

Baseline

9

175

Healthy

10

560

Baseline

10

481

Healthy


In 2 to 3 sentences in a Microsoft Word document, answer the following questions:

  • What is the independent variable in this study?
  • What are the levels of that independent variable?
  • What is the dependent variable?

In: Math

6. It's presidential primary season and canvassers are out talking to voters. In a city that's...

6. It's presidential primary season and canvassers are out talking to voters. In a city that's known to have 20% registered republicans, 5 canvassers each go to 50 randomly selected homes to ask about voting preferences. Amazingly, every home has someone willing to talk. Which of the following is the most plausible sequence of republican voters met by these canvassers?

a. 5%, 80%, 65%, 8%, 70%

b. they're all equally plausible

c. 15%, 25%, 22%, 28%, 20%

d. 20%, 20%, 20%, 20%, 20%

7.

The average birthweight of babies in Oregon is 3500 grams with a standard deviation of 500 grams. You collected 100 samples of 100 babies and calculated the mean weight of each of the samples. You then graph the means that you've calculated. What does your distribution of sample means look like?

a. it's pretty normal

b. it has no particular shape

c. it's skewed either left or right

8.

Still thinking about your distribution of the samples of baby weights, what would the standard deviation of your 100 sample means be?

a. more than 500 grams

b. not enough information to know

c, less than 500 grams

d. same as the population, 500 grams

In: Math

Hypothesis testing terminology a. Level of Significance d. Power g. Test Statistic b. Alternative Hypothesis e....

Hypothesis testing terminology

a. Level of Significance d. Power g. Test Statistic
b. Alternative Hypothesis e. Effect Size h. Type I Error
c. Null Hypothesis f. Type II Error i. Significant Effect

Match each of the following descriptions with its corresponding term from the list above. Enter the letter corresponding to the correct term in the blank.

A mistake researchers can make when they don’t conclude, for example, that a treatment has an effect when it actually does
The hypothesis for a hypothesis test that predicts that the independent variable has an effect
The probability of rejecting the null hypothesis when it is false
A treatment has this if the decision from the hypothesis test is to reject the null hypothesis
An indication of the magnitude of the treatment effect
A value computed using sample data that is used to decide whether to reject the null hypothesis
A mistake researchers can make when they conclude, for example, that a treatment has an effect when it does not
The hypothesis for a hypothesis test that predicts that the independent variable has no effect
The maximum probability the researcher is willing to accept of making a Type I error

In: Math

Examples, please Paired design with repeated measures Paired design with matched pairs

Examples, please

Paired design with repeated measures

Paired design with matched pairs

In: Math

At a recent halloween party, the women appeared to be consuming more packages of halloween candy...

At a recent halloween party, the women appeared to be consuming more packages of halloween candy than were the men. If the mean number of packages consumed by the 3 men was 4, and that for the 7 women was 6, and the standard deviation for the whole group was 2 packages, what was the correlation between gender and the number of packages consumed?

I am having difficulties understanding how to solve this homework problem.

In: Math

Independent random samples of professional football and basketball players gave the following information. Assume that the...

Independent random samples of professional football and basketball players gave the following information. Assume that the weight distributions are mound-shaped and symmetric.

Weights (in lb) of pro football players: x1; n1 = 21

249 261 255 251 244 276 240 265 257 252 282
256 250 264 270 275 245 275 253 265 271

Weights (in lb) of pro basketball players: x2; n2 = 19

202 200 220 210 193 215 221 216 228 207
225 208 195 191 207 196 181 193 201

(a) Use a calculator with mean and standard deviation keys to calculate x1, s1, x2, and s2. (Round your answers to one decimal place.)

x1 =
s1 =
x2 =
s2 =


(b) Let μ1 be the population mean for x1 and let μ2 be the population mean for x2. Find a 99% confidence interval for μ1μ2. (Round your answers to one decimal place.)

lower limit    
upper limit    


(c) Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the 99% level of confidence, do professional football players tend to have a higher population mean weight than professional basketball players?

Because the interval contains only negative numbers, we can say that professional football players have a lower mean weight than professional basketball players.Because the interval contains both positive and negative numbers, we cannot say that professional football players have a higher mean weight than professional basketball players.    Because the interval contains only positive numbers, we can say that professional football players have a higher mean weight than professional basketball players.


(d) Which distribution did you use? Why?

The standard normal distribution was used because σ1 and σ2 are known.The standard normal distribution was used because σ1 and σ2 are unknown.    The Student's t-distribution was used because σ1 and σ2 are unknown.The Student's t-distribution was used because σ1 and σ2 are known.

In: Math

Real Fruit Juice (Raw Data, Software Required): A 32 ounce can of a popular fruit drink...

Real Fruit Juice (Raw Data, Software Required):
A 32 ounce can of a popular fruit drink claims to contain 20% real fruit juice. Since this is a 32 ounce can, they are actually claiming that the can contains 6.4 ounces of real fruit juice. The consumer protection agency samples 30 such cans of this fruit drink. The amount of real fruit juice in each can is given in the table below. Test the claim that the mean amount of real fruit juice in all 32 ounce cans is 6.4 ounces. Test this claim at the 0.01 significance level.

(a) What type of test is this?

This is a left-tailed test.

This is a two-tailed test.    

This is a right-tailed test.


(b) What is the test statistic? Round your answer to 2 decimal places.
t-x= ?

(c) Use software to get the P-value of the test statistic. Round to 4 decimal places.
P-value = ?

(d) What is the conclusion regarding the null hypothesis?

reject H0

fail to reject H0    


(e) Choose the appropriate concluding statement.

There is enough data to justify rejection of the claim that the mean amount of real fruit juice in all 32 ounce cans is 6.4 ounces.

There is not enough data to justify rejection of the claim that the mean amount of real fruit juice in all 32 ounce cans is 6.4 ounces.    

We have proven that the mean amount of real fruit juice in all 32 ounce cans is 6.4 ounces.

We have proven that the mean amount of real fruit juice in all 32 ounce cans is not 6.4 ounces.

    
    DATA ( n = 30 )
Real Juice
ounces   
6.32
6.55
6.56
6.39
6.27
6.02
6.29
6.06
6.11
6.05
6.51
6.02
6.51
6.01
6.65
6.34
6.54
6.46
5.97
6.12
6.23
6.27
5.80
6.59
5.72
6.71
6.46
6.76
6.32
6.26

In: Math

According to the National Association of Colleges and Employers, the 2015 mean starting salary for new...

According to the National Association of Colleges and Employers, the 2015 mean starting salary for new college graduates in health sciences was $51,541. The mean 2015 starting salary for new college graduates in business was $53,901. † Assume that starting salaries are normally distributed and that the standard deviation for starting salaries for new college graduates in health sciences is $11,000. Assume that the standard deviation for starting salaries for new college graduates in business is $17,000.

(a) What is the probability that a new college graduate in business will earn a starting salary of at least $61,000? (Round your answer to four decimal places.)

(b) What is the probability that a new college graduate in health sciences will earn a starting salary of at least $61,000? (Round your answer to four decimal places.)

(c) What is the probability that a new college graduate in health sciences will earn a starting salary less than $42,000? (Round your answer to four decimal places.)

(d) How much would a new college graduate in business have to earn in dollars in order to have a starting salary higher than 99% of all starting salaries of new college graduates in the health sciences? (Round your answer to the nearest whole number.)

In: Math

Seasonal affective disorder (SAD) is a type of depression during seasons with less daylight (e.g., winter...

Seasonal affective disorder (SAD) is a type of depression during seasons with less daylight (e.g., winter months). One therapy for SAD is phototherapy, which is increased exposure to light used to improve mood. A researcher tests this therapy by exposing a sample of patients with SAD to different intensities of light (low, medium, high) in a light box, either in the morning or at night (these are the times thought to be most effective for light therapy). All participants rated their mood following this therapy on a scale from 1 (poor mood) to 9 (improved mood). The hypothetical results are given in the following table.

Light Intensity
Low Medium High
Time of
Day
Morning 5 5 7
6 5 8
4 4 6
7 7 9
5 9 5
6 7 8
Night 5 6 9
7 8 7
6 7 6
7 5 7
4 9 7
3 8 6

(a) Complete the F-table and make a decision to retain or reject the null hypothesis for each hypothesis test. (Round your answers to two decimal places. Assume experimentwise alpha equal to 0.05.)

Source of
Variation
SS df MS F
Time of day
Intensity
Time of
day ×
Intensity
Error
Total


State the decision for the main effect of the time of day.

Retain the null hypothesis or Reject the null hypothesis.    


State the decision for the main effect of intensity.

Retain the null hypothesis or Reject the null hypothesis.    


State the decision for the interaction effect.

Retain the null hypothesis or Reject the null hypothesis.    


(b) Compute Tukey's HSD to analyze the significant main effect.

The critical value is_________ for each pairwise comparison.


Summarize the results for this test using APA format.

In: Math

Annual starting salaries for college graduates with degrees in business administration are generally expected to be...

Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $20,000 and $50,000 . Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired.

a. What is the planning value for the population standard deviation? _____

b. How large a sample should be taken if the desired margin of error is $300? Round your answer to next whole number.

____________

$250? _______

$140? _______

c. Would you recommend trying to obtain the $140 margin of error? Explain.
- Select your answer -Yes, it always better to be more accurate.No, the sample size would probably be too time consuming and costly.Item 5

In: Math

how can you apply the concept "testing a claim about a proportion" to the following activities:...

how can you apply the concept "testing a claim about a proportion" to the following activities:

watching football

working at a gym

what can you test in these two activites and how can we apply the concept

In: Math

Statistical estimation A metalworking company has 192 operators. In a random sample of 50 of them,...

Statistical estimation

A metalworking company has 192 operators. In a random sample of 50 of them, the average number of overtime hours worked in a week and their standard deviation was 9.6 and 6.2 hours respectively.

a.In order to program the economic resources for all the personnel of operators, the personnel department decides: To estimate with a confidence of 96% the average number of overtime hours worked by each operator during a week.
b. Estimate with a 99% confidence the total number of overtime hours worked by the company's operators for a week.

In: Math