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 (μ1 − μ2 > 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 =
(x1 − x2) − δ | ||||||
|
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 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 (μ1 − μ2 > 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 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. |
|
Additional Materials
In: Statistics and Probability
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.
In: Statistics and Probability
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 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, 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 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 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 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
|
In: Statistics and Probability
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.
p̂ =
(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 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 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 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 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.
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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.
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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