Questions
Register Balance: Here we investigate whether the register balance at a local retail store is better...

Register Balance: Here we investigate whether the register balance at a local retail store is better when a manager is on-duty compared to when a manager is off-duty. Evidence like this might be used to determine whether or not an employee is stealing money from the register when no manager is around. The table below gives the register balance (0 means the register balance is right on, negative means there is less money than there should be, and positive means there is more money than there should be) for 10 days when the manager is present and for 10 days when the manager is not present. Test the claim that the mean register balance for all days when the manager is on duty is greater than the mean register balance when the manager is off duty. Test this claim at the 0.01 significance level.

Manager On Duty No Manager
count Register Balance (x1) Register Balance (x2)
1 -5 2
2 1 -8
3 -7 -15
4 -4 -10
5 5 -10
6 -3 0
7 -2 -12
8 -1 -5
9 -7 0
10 -5 -14
x -2.80 -7.20
s2 13.96 37.73
s 3.74 6.14

If you are using software, you should be able copy and paste the data directly into your software program.

(a) The claim is that the difference in population means is positive (μ1μ2 > 0). 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) Use software to calculate the test statistic or use the formula
t =

(x1x2) − δ
s12
n1
+
s22
n2

where δ is the hypothesized difference in means from the null hypothesis. Round your answer to 2 decimal places.

t =


To account for hand calculations -vs- software, your answer must be within 0.01 of the true answer.

(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 H0fail to reject H0    


(e) Choose the appropriate concluding statement.

The data supports the claim that the mean register balance for all days when the manager is on duty is greater than the mean register balance when the manager is off duty.There is not enough data to support the claim that the mean register balance for all days when the manager is on duty is greater than the mean register balance when the manager is off duty.    We reject the claim that the mean register balance for all days when the manager is on duty is greater than the mean register balance when the manager is off duty.We have proven that someone is stealing money from the register when the manager is not on duty.

In: Statistics and Probability

Register Balance (Raw Data, Software Required): Here we investigate whether the register balance at a local...

Register Balance (Raw Data, Software Required):
Here we investigate whether the register balance at a local retail store is better when a manager is on-duty compared to when a manager is off-duty. Evidence like this might be used to determine whether or not an employee is stealing money from the register when no manager is around. The table below gives the register balance (0 means the register balance is right on, negative means there is less money than there should be, and positive means there is more money than there should be) for 10 days when the manager is present and for 10 days when the manager is not present. Test the claim that the mean register balance for all days when the manager is on duty is greater than the mean register balance when the manager is off duty. Test this claim at the 0.01 significance level.

Manager On Duty No Manager
count Register Balance (x1) Register Balance (x2)
1 -4 2
2 1 -9
3 -6 -15
4 -3 -10
5 5 -10
6 -3 0
7 -2 -12
8 -1 -5
9 -7 0
10 -5 -14

You should be able copy and paste the data directly into your software program.

(a) The claim is that the difference in population means is positive (μ1μ2 > 0). What type of test is this?

This is a right-tailed test.This is a two-tailed test.    This is a left-tailed test.


(b) Use software to calculate the test statistic. Do not 'pool' the variance. This means you do not assume equal variances.
Round your answer to 2 decimal places.

t =



(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 H0fail to reject H0    


(e) Choose the appropriate concluding statement.

The data supports the claim that the mean register balance for all days when the manager is on duty is greater than the mean register balance when the manager is off duty.There is not enough data to support the claim that the mean register balance for all days when the manager is on duty is greater than the mean register balance when the manager is off duty.    We reject the claim that the mean register balance for all days when the manager is on duty is greater than the mean register balance when the manager is off duty.We have proven that someone is stealing money from the register when the manager is not on duty.

In: Statistics and Probability

AM -vs- PM Height (Raw Data, Software Required): It is widely accepted that people are a...

AM -vs- PM Height (Raw Data, Software Required):
It is widely accepted that people are a little taller in the morning than at night. Here we perform a test on how big the difference is. In a sample of 30 adults, the morning height and evening height are given in millimeters (mm) in the table below. Use this data to test the claim that, on average, people are more than 10 mm taller in the morning than at night. Test this claim at the 0.01 significance level.



(a) The claim is that the mean difference (x - y) is more than 10 mm (μd > 10). What type of test is this?

This is a two-tailed test.This is a right-tailed test.    This is a left-tailed test.


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

d

=

(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 H0fail to reject H0    


(e) Choose the appropriate concluding statement.

The data supports the claim that, on average, people are more than 10 mm taller in the morning than at night.There is not enough data to support the claim that, on average, people are more than 10 mm taller in the morning than at night.    We reject the claim that, on average, people are more than 10 mm taller in the morning than at night.We have proven that, on average, people are more than 10 mm taller in the morning than at night.

    
    
AM Height (x) PM Height (y) (x - y)
1584 1575 9
1429 1416 13
1502 1494 8
1477 1468 9
1766 1758 8
1598 1586 12
1490 1478 12
1768 1755 13
1696 1684 12
1557 1545 12
1432 1418 14
1753 1743 10
1786 1772 14
1719 1704 15
1752 1742 10
1683 1667 16
1598 1585 13
1714 1696 18
1459 1449 10
1683 1674 9
1685 1676 9
1484 1472 12
1762 1754 8
1481 1472 9
1656 1649 7
1669 1660 9
1542 1529 13
1585 1579 6
1728 1719 9
1571 1560 11

Additional Materials

In: Statistics and Probability

4th Grade (Class 1) 4th Grade (Class 2) 12 10 15 12 21 16 21 17...

4th Grade (Class 1) 4th Grade (Class 2)
12 10
15 12
21 16
21 17
22 17
22 19
22 19
25 22
26 22
27 22
27 27
31 28
32 29
33 29
33 31
36 31
37 31
38 33
41 33
43 37
44 39
45 43
45 43
47 47
55 49
57 57

The collected data is from two 4th grade (All female classes - Age 10) Fitnessgram pacer tests. Once you have analyzed the data, please answer the following questions.

  1. What was the mean, median, and mode for each class?

In: Statistics and Probability

Table 6-3. Calculation of MAD Time Period      Actual Sales      3 Month Moving Average          Absolute Deviation     

Table 6-3. Calculation of MAD

Time Period      Actual Sales      3 Month Moving Average          Absolute Deviation        Weighted Moving Average

W=.1,.3,.6         Absolute Deviation        Exponential Smoothing Ft = 350 Absolute Deviation

1          230                                                      350      

2          238                                                      326      

3          260                                                      308      

4          275      243                  250                  299      

5          300      258                  267                  294      

6          285      278                  289                  295      

7          270      287                  289                  293      

8          290      285                  278                  289      

9          305      282                  284                  289      

10        320      288                  297                  292      

11        335      305                  313                  298      

12                    320                  328                  305      

SUM =  

                        n =                                                       

                        MAD   

73) Refer to Table 6-3. Calculate MAD for all forecasting models. The model you would use for your forecast, based on MAD is _____________________________________.

Please show work.  

In: Statistics and Probability

Let the shape of the distribution of the time required to get an oil change at...

Let the shape of the distribution of the time required to get an oil change at a 20-minute oil-change facility is unknown. However, records indicate that the mean time is 21.6 minutes, and the standard deviation is 4.5 minutes. Round to four decimal points, as needed. Include Excel formulas. *

What is the probability that a random sample of 40 oil changes results in a sample mean time of less than 20 minutes? *

What is the mean time required to get an oil change when the probability that it takes longer than that time is 5%? *

In: Statistics and Probability

III. Proof of assertion regarding a mean: σ unknown 1. In a promotions contest in Ireland,...

III. Proof of assertion regarding a mean: σ unknown

1. In a promotions contest in Ireland, the ages of unsuccessful applicants and those who succeeded in obtaining the promotion were studied. For unsuccessful applicants, a sample of 23 was used, which reflected an average age of 47 and a standard deviation of 7.2. The sample of them succeeded, showed an average of 43.9, with a standard deviation of 5.9, in 30 cases analyzed. It uses a .05 level of significance to test the assertion that unsuccessful applicants come from a larger average population than that of successful applicants.

In: Statistics and Probability

Carleton Chemical claims that they can produce more than 800 tons of benzene on average per...

Carleton Chemical claims that they can produce more than 800 tons of benzene on average per week. A random sample of 36 weeks of production yields a mean and SD as follows: x ̅=823 and s = 79.8. At the = .05 significance level, does the sample data provide significant evidence to support the claim made by the Carleton Company?

a) Set up hypotheses Ho and Ha to test the company’s claim.

Go to STAT >> TESTS, and choose T-test – select the STATS option and enter the given sample information.

b) State the test statistic and p-value given by the calculator.

c) Compare the p-value to  = .05. State the appropriate decision (either Reject Ho or Fail to reject Ho), along with an explanation.

d) Interpret this decision to write a conclusion for the test.

In: Statistics and Probability

A random sample of 18 Belgium chocolate bars has, on average, 345 calories per bar, with...

A random sample of 18 Belgium chocolate bars has, on average, 345 calories per bar, with a standard deviation of 28 calories.  Construct a 95% confidence interval for the true mean calorie content of these chocolate bars.  Assume that the distribution of the calorie content is approximately normal

In: Statistics and Probability

Salmon (Raw Data, Software Required): Assume that the weights of Chinook Salmon in the Columbia River...

Salmon (Raw Data, Software Required):
Assume that the weights of Chinook Salmon in the Columbia River are normally distributed. You randomly catch and weigh 15 such salmon. The data is found in the table below. Test the claim that the mean weight of Columbia River salmon is greater than 26 pounds. Test this claim at the 0.05 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 H0fail to reject H0    


(e) Choose the appropriate concluding statement.

The data supports the claim that the mean weight of Columbia River salmon is greater than 26 pounds.There is not enough data to support the claim that the mean weight of Columbia River salmon is greater than 26 pounds.    We reject the claim that the mean weight of Columbia River salmon is greater than 26 pounds.We have proven that the mean weight of Columbia River salmon is greater than 26 pounds.

    
    DATA ( n = 15 )
Salmon Weights
Pounds   
25.7
24.8
36.0
26.6
18.9
22.5
33.2
25.8
23.5
26.8
32.5
27.7
28.8
28.3
31.5

In: Statistics and Probability

Corn: In a random sample of 120 ears of corn, farmer Carl finds that 18 of...

Corn: In a random sample of 120 ears of corn, farmer Carl finds that 18 of them have worms. Carl claims that less than 20% of all his corn has worms. Test this claim at the 0.05 significance level.

(a) What is the sample proportion of corn with worms? Round your answer to 3 decimal places.

=



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

z =



(c) What is the P-value of the test statistic? Round your answer to 4 decimal places.
P-value =


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

reject H0fail to reject H0    


(e) Choose the appropriate concluding statement.

The data supports the claim that less than 20% of Carl's corn has worms.There is not enough data to support the claim that less than 20% of Carl's corn has worms.     We reject the claim that less than 20% of Carl's corn has worms.We have proven that less than 20% of Carl's corn has worms.

In: Statistics and Probability

Problem: Susan Sound predicts that students will learn most effectively with a constant background sound, as...

Problem: Susan Sound predicts that students will learn most effectively with a constant background sound, as opposed to an unpredictable sound or no sound at all. She randomly divides twenty-four students into three groups of eight. 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. Those in group 3 study with no sound at all. After studying, all students take a 10 point multiple choice test over the material. Their scores follow:
group test scores


1) constant sound 7 4 6 8 6 6 2 9
2) random sound 5 5 3 4 4 7 2 2
3) no sound 2 4 7 1 2 1 5 5

Please give me step by step how to answer the question.

In: Statistics and Probability

iPhone 6 battery life ~ The life of a fully charged iPhone 6 is normally distributed...

iPhone 6 battery life ~ The life of a fully charged iPhone 6 is normally distributed with an average of 13 hours and a standard deviation of 1.5 hours.

What is the life of a fully charged iPhone 6 battery for the 20th percentile? Enter your answer to 4 decimal places.

Your Answer:

Question 1 options:

Answer

In: Statistics and Probability

According to a recent report, 68% of Internet searches in a particular month used the Google...

According to a recent report, 68% of Internet searches in a particular month used the Google search engine. Assume that a sample of 21 searches is studied. Round the answers to four decimal places.

(a) What is the probability that exactly 17 of them used Google?

(b) What is the probability that 12 or fewer used Google?

(c) What is the probability that more than 17 of them used Google?

(d) Would it be unusual if fewer than 9 used Google?

It ▼(Choose one) be unusual if fewer than 9used Google since the probability is _. .

In: Statistics and Probability

This quiz covers the first half of Section 7.1 (constructing confidence intervals using t statistics) and...

This quiz covers the first half of Section 7.1 (constructing confidence intervals using t statistics) and is due at 9:00 PM on Thursday, April 2.

You will have two attempts at this quiz. I suggest you do all of them first on paper. After the first attempt, you should write down which ones you answered correctly and which one incorrectly, as well as any hints that were given. Then re-solve the problems and take the quiz a second time. Note that the responses will appear in a different order the second time you take the quiz. Correct answers will appear on Canvas, twenty four hours after the quiz is due.

By submitting this quiz you affirm you did not receive any help from anyone in completing the quiz, except for clarifications provided by the instructor.  Using the videos, textbook and handbook is acceptable.

Question 11 pts

William Gosset worked in quality control for a major beverage company. He developed the t-distribution to:

Group of answer choices

monitor the sugar content of Pretty Things craft beer.

improve the consistency of Guinness ale.

determine if drinking Gallo's Zinfandel red wine decreased heart attack risk,

compare the richness of Nescafe regular and decaf coffees.

Flag this Question

Question 21 pts

A t distribution is used in constructing confidence intervals for all the following reasons EXCEPT:

Group of answer choices

To account for the extra uncertainty when the population standard deviation (sigma) is unknown.

When the sample size is too large to use z statistics.

When we use the sample standard deviation (s) in constructing the margin of error.

When we do not have any historical data that can help us make our estimate.

Flag this Question

Question 31 pts

When computing the margin of error (MOE) using a t-distribution, the following are all true EXCEPT:

Group of answer choices

The MOE increases when t* is used in place of z*.

The MOE increases when the degrees of freedom (df) increases.

The MOE increases when the standard error of the mean increases.

The MOE increases when a higher confidence level is used.

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Question 41 pts

Questions 4 through 7 are the steps required to solve Exercise #7.30(a) on IPS p. 428. This study investigates the hours per month spent on the Internet by U.S. residents, age 18 to 24. The sample results are: n= 75, x-bar=28.5, and s=23.1 . We are to construct a 95% confidence interval based on this information.

What is critical value of t (that is, t*) for this confidence interval?

Group of answer choices

1.960

1.990

2.000

2.009

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Question 51 pts

Continuing Question 4, the standard error of the mean (SEM) is:

Group of answer choices

.3080

23.10

2.685

2.667

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Question 61 pts

Rounded to one decimal place, as x-bar was, the margin of error (MOE) for your confidence interval will be:

Group of answer choices

4.7

5.9

5.6

5.3

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Question 71 pts

Which of the following conclusions is CORRECT?

Group of answer choices

We are 95% sure the mean hours spent per month on the Internet by college students is between 23.2 and 33.8.

We are 95% sure the mean hours spent per month on the Internet by U.S. residents, age 18 to 24, is between 23.5 and 34.1.

We are 95% sure the mean hours spent per month on the Internet by U.S. residents, age 18 to 24, is between 23.2 and 33.8.

We are 95% sure the mean hours spent per week on the Internet by U.S. residents, age 18 to 24, is between 23.5 and 34.1.

In: Statistics and Probability