Stats test
.02921709 statistical interpretation? What is the conclusion?
Since the P-value is very small, we are confident that the slope is indeed positive. We can conclude that the fliers did influence the number of customers.
A company is testing a new type of engine lubricant and their claim is that this new product (by lowering the frictions), would increase the car gas mileage. Upload Lubricants and perform the appropriate test and comment on the result.
Since we want to show that the new product increases the gas mileage, a one-sided t-test seems appropriate
.001963 Statistical interpretation? What is the conclusion?
Statistical interpretation : Since the P-value is very small we can conclude that one of the two averages is significantly higher. Conclusion: We can be confident that the new lubricant increases the gas mileage.
P value .0058 Statistical interpretation? What is the conclusion?
Statistical interpretation: Since the P-value is too large we cannot statistically conclude that the averages are different. Conclusion: We do not have enough statistical evidence to conclude that the drug lowered cholesterol level.
P value .00006 statistical interpretation? Conclusion?
Statistical interpretation: Since the P-value is very small, we can confidently claim the salaries are different. Conclusion: The salaries for Nursing and Pre Law majors are indeed different.
t-test
TTest(Array1,Array2,2,3)
Confidence interval:
We need one data set and the question if the average of this data set is different (larger/smaller) than a fixed number
T-test:
We need two variables (two data sets) and the question if the average of one variable (i.e. X) is different (larger/smaller) than the average of variable Y. If the question is about the difference of averages we use two-sided T-test, otherwise we use one-sided T-test.
Regression:
We need two variables (two data sets) and the question if variable X is influencing variable Y.
If the P-value is smaller than 5% (sometimes 1%)
We say: We are confident that the averages tested are significantly different.
If P-value is larger than 5% (sometimes 1%)
We say: We do not have enough evidence to statistically differentiate these two averages.
In order to increase the number of customers your boss decided to try distributing fliers. The data represent the number of fliers as well as the number of customers per given day. Given the data, your job is to perform the appropriate statistical test or procedure and help your boss with the decision.
We use regression. Why? Because the manager is interested if variable X (fliers) is influencing variable Y (the customers). Why not use a t-test? It is true that we have two data sets, but the manager is not asking if the average number of fliers is different from the number of customers (Remember: T-test is used when one compares two averages).
Rule of Thumb for the Confidence Interval Procedure
When to use it? If you are asked to compare the average of ONE data set to ONE particular number. Let us call this number A. How to argue? Once you produce the confidence interval you have three options:
It is believed that the more math credits students take during their college years, the higher their salary will be. Given the data your job is to confirm or disprove this assertion.
regression
It is reasonable to assume that IQ and weight of a person should not be related. Given the data, your job is to check if this assertion is indeed reasonable or not.
regression
Test if the more assaults occur in New York the more murders are expected in New York
regression
Using data on 33 second-year students enrolled in a Statistics course, test if more visits to open lab sessions lead to higher final grades.
regression
test if older houses are cheaper to buy
regression
there seems to be a connection between students SAT scores and their College GPA. Perform appropriate test or statistics procedure to address this assertion.
regression
to exteend formula down
select all ctrl-d
p value= .220243 s. i? conclusion?
the p value is too large to have a conclusive answer we cannot confirm
The data represents a fictional case where a new cholesterol lowering drug was tested. Column A represents 20 patients and their cholesterol levels after treatment while column B represents their cholesterol levels before the treatment. Our job is to decide if the drug worked. We will follow our usual template:
ttest one sided since it is called lowered (change would be two sided )
Test if there is a significant difference between men's and women's wages.
two sided
Do Nursing and Pre Law majors have different salaries? Given the data answer this question.
two sided t-test
P- Value .513 What is the statistical interpretation? What is the conclusion?
1. Since the P-value 0.513 is larger than 5% we do not have enough statistical evidence to distinguish these two averages. 2. Although the college environment and student-teacher relationships are very different from high school, college grading does not seem to be different
p-value: .2 What is the statistical interpretation? What is the conclusion?
1. Since the P-value is larger than 5%, we cannot statistically conclude that the averages are different. 2. Although the averages are very different, the statistical test could not conclusively claim that they are significantly different.
P value is .002738 What is the statistical interpretation? What is the conclusion?
1. Since the P-value is much smaller than 5%, we are very confident that the average prices are indeed different. 2. We can confirm that the prices of houses with one bathroom, versus houses with more than one bathroom, are significantly different.
st dev= 14.11 alpha= .05 data points= 25 what is the confidence interval? Write a one-line statistical interpretation. Write a one-line conclusion.
=Confidence.t(alpha, standard dev, size) = confidence.t(.05,14.11,25) 5.824 Since the confidence interval is below the old average of 45 min we can (with 95% confidence) claim that the new average is below 45 min. Yes, we can confidently claim that the new procedure lowered the average waiting time.
p value .0016 statistical interpretation? Conclusion?
Since it is very low we can confidently claim that variable X influences variable Y
regression
Checks if one variable influences the other variable (if related) Key words: connected, related, the higher the variable X, the higher/lower the variable Y, etc... Data Data analysis Regression
One sided ttest
Compares two averages to see if one is higher (or lower) than the other one. **Note: Check the averages first to see if they agree with the claim! If they do NOT agree with the claim, no need to compute the P-value. ** (Say "none of these" for the p-value question)** = TTest (array 1 , array 2, 1, 3)
Two sided ttest
Compares two averages to see if they are different Keywords/phrases: any changes in averages any difference in averages etc... =TTest (array 1, array 2, 2 , 3)
The manufacturer introduced a new version of the machine and was sued for not disclosing the additional maintenance cost. The average annual maintenance cost for the older model was $1070. Given the data for the recent machines decide if the lawsuit holds water. Use a 5% P-value What test/procedure did you perform? What is the P-value/margin of error? Write a one-line statistical interpretation. Write a one-line conclusion.
Confidence Interval What test/procedure did you perform? Correct answer: Confidence interval. What is the P-value/margin of error? Correct answer: Margin of error = $252.87. Write a one-line statistical interpretation. Correct answer: Since the confidence interval [1003.57, 1509.32] contains the company's claim of $1070, we cannot statistically claim that new machine maintenance cost is different from $1070. Write a one-line conclusion. Correct answer: The lawsuit does not have merit.
test if the month of may has more than 130 tornadoes?
Confidence interval
There were 13000 inspected files. The percentage of delinquency was 4.3% and the standard deviation was 0.2. Since Frank is risking his job, he wants to be very (very) certain, so we will work with 99.9% certainty. Can you help Frank? What test/procedure should be performed? What is the margin of error? What is the statistical interpretation? What is the conclusion?
Confidence interval. Why? Because Frank needs to compare the average of his data (i.e. 4.3%) with the required fixed number (in this case 5%). What is the margin of error? st dev= .2 data size: 13000 alpha: 0.001 confidence level= .005773 What is the statistical interpretation? (.043-.00577)(.043+.00577) 3.72% 4.87% What is the conclusion? Frank should stop worrying. It is very likely that the percentage of delinquent files is less than 5%.
This data represent a fictitious survey of 100 patients that used a particular medicine. The number "1" stands for patients that reported a side effect (drowsiness), while "0" stands for patients that did not report any drowsiness as a side effect. In order to be sold without a doctor's prescription, the US Federal Drug Administration (FDA) requires that a drug must have less than 10% of patients reporting drowsiness as a side effect. The FDA also requires at least 99% certainty for statistical testing purposes. What test/procedure should be performed? What is the margin of error? What is the statistical interpretation? What is the conclusion?
Confidence interval. Why? Because we need to compare the average of our data with the required fixed number (in this case required number is given by FDA). We need to use the standard deviation (type = StDev(Array). The result is 0.27. The data size = 100 and alpha = 0.01 (since FDA requires 99% certainty). Thus the margin of error = CONFIDENCE.T(0.01, 0.27, 100) = 0.070913 The average is 0.08 which means that the 99% confidence interval is [0.08 - 0.07, 0.08 + 0.07] = [1%, 15%]. Since the FDA requirement of 10% is inside this interval we cannot (with 99% certainty) claim that this new drug has a percentage of drowsiness below 10%. Although the percentage of people that experience drowsiness is 8% (which is less than FDA required 10%), we cannot statistically conclude that this drug has less than 10% of patients that experience drowsiness
The P-value = 0.245782168. Write a one-line statistical interpretation. Write a one-line conclusion.
Correct answer: Since the P-value is too large we cannot conclude that there is any significant difference between the two averages. Correct answer: The presence of soda machines does not seem to change the average weight of the students.
r^2: .2804 pvalue: .000154 Is this a very strong connection?
Since the P-Value is small, we are very certain that there is a connection between murders in Texas and Vermont. No. Because R-squared is not very high.
A hospital is trying to implement a new patient check-in procedure. It is known, that for this particular hospital, the average waiting time for patients is 45 min. Your job is to determine if this new procedure provides significantly shorter waiting time. confidence interval
In our case the Hospital data in the above Task, we type = Average(Array1) and get 34.24. And for standard deviation we type = StDev(Array1) and the result is 14.11. To do so just type = CONFIDENCE.T(alpha, standard dev, size). In our case alpha will be 0.05, standard dev = 14.11 and size refers to the number of data points (25 data points). Thus type = CONFIDENCE.T(0.05, 14.11, 25) and the result is 5.824. By letting alpha = 5%, we are constructing 95% confidence interval (100% - alpha). Alpha has a role similar to the P-value. Interval = [Average - Margin of error, Average + Margin of Error]; in our case [34.24 - 5.82, 34.24 + 5.82] = [28.42, 40.06]. Statistical Interpretation: Since the confidence interval is below the old average of 45 min, we can (with 95% confidence) claim that the new average is below 45 min. Conclusion: Yes, we can confidently claim that the new procedure lowered the average waiting time. In a nutshell: The "confidence interval" tells us that the average waiting time for new check-in procedure is somewhere between 28.42 min and 40.06 min. And we can be 95% confident that this is correct. Comment: In most statistical applications we are satisfied with a 95% confidence interval (i.e. alpha = 0.05). However, one can always compute 99% or 99.9% confidence interval; just let alpha = 0.01 or 0.001.
The data represent weekly profit (in thousands) for 20 restaurants. Ten restaurants were equipped with new TV sets while the remaining ten were not. Management wants to see if the new TV sets increased the profit. Given the data, your job is to perform the appropriate statistical test or procedure and help management with their decision what is the p value? Write a one-line statistical interpretation. Write a one-line conclusion.
The appropriate test would be a one-sided t-test. Why? Because here we are asked to compare two averages (thus the t-test) and moreover, we are asked to check if one is larger than the other (thus the one-sided test). This question is not necessary since P-value should NOT be computed. Why? Before performing the one-sided T-test, we must first check if the actual averages are in agreement with the claim. Namely, the average profit for restaurants with new TVs is 7.5 which is lower than for the restaurants with no new TV (there the average is 9.5). Hence, if one-sided T-test is selected and averages are not in agreement with the assumption, then no P-value should be computed. Test is inconclusive. Why? Since the average profit for restaurants with new TVs, 7.5, is lower than the average profit for the restaurants with no new TV, 9.5, we cannot conclude that new TVs improved profit.
P value 2.415 E-08 Statistical interpretation? What is the conclusion?
The statistical interpretation: Since the P-value is very small we are very confident that there is a relationship between SAT and GPA score. Conclusion: Clearly, we can conclude that, on average, the higher the students' SAT score the higher their GPAs will be.
Two production plants were examined and the number of work-related accidents was reported on a daily basis. The data is presented below. The following was performed in Excel "=TTest(I2:I12, J2:J12, 2, 3)." And the number returned by Excel was 0.411. Write a one-line explanation about what was exactly tested. Write a one-line precise statistical interpretation commenting on P-value. Write one to two lines conclusion explaining the P-value in more layman terms
Write a one-line explanation about what was exactly tested. Your answer: 0 Correct answer: Since we used parameters "2, 3," this was a two-sided t-test and we tested if the two averages (for two factories) are different from each other. Write a one-line precise statistical interpretation commenting on P-value. Write a one-line precise statistical interpretation commenting on P-value. Correct answer: Since the P-value 0.411 is too large, we cannot conclude that the averages are different. Write one to two lines conclusion explaining the P-value in more layman terms. Your answer: 000 Correct answer: In conclusion, we cannot statistically differentiate the two factories.
p value 1.3694e-06 statistical interpretation? Conclusion?
a. The P-value is much smaller than 5% thus we are very certain that the averages are significantly different. b. Statistical interpretation agrees with the intuition that men earn more than women on average.
P value= none statistical intrpretation? conclusion?
average data is inconsistent with the claim we cannot claim florida has more per year on average
basic histogram curves
bell shape, double bell shape, skewed
A hospital is trying to implement a new patient check-in procedure. It is known, that for this particular hospital, the average waiting time for patients is 45 min. Your job is to determine if this new procedure provides significantly shorter waiting time.55
confidence interval
Is it reasonable to claim that people who took more than 10 math credits will have an average salary of $43000 or more?
confidence interval
Is it reasonable to claim that students with 8 or more visits to open lab sessions have average final grade greater than 78?
confidence interval
Which of the following is false? (4 points) a. We use two sided t-test when we are checking if the averages of the two variables are different, with no prior assertion on which is larger of the two. b. We use one sided t-test when we are checking if the averages of the two variables are different, but now we are more specific. We want to show that X has a significantly larger/smaller average than Y. c. We need to perform the averages of the two variables data set when one sided t-test is chosen in order to see if the averages are consistent with the claim. d. The one sided t-test and two sided t-test performed on the same data sets should return the same P-value. e. The P-value excel returns on a one sided t-test is incorrect and meaningless if the averages computed on the data sets are inconsistent with the claim
d. The one sided t-test and two sided t-test performed on the same data sets should return the same P-value.
probability
frequency/max points
Test if Florida has more robberies than assaults on average. Answer questions 6 to 9. (Pick the closest answer)
one sided
Test if West Virginia's vehicle thefts are lower than Nebraska's vehicle thefts on average.
one sided
Test if unmarried women's wages is greater than married women's wages on averag
one sided
Using data on 90 second-year students enrolled in a Statistics course, test if students who are assiduous in attending lectures have higher final grades than students who are usually absent
one sided
