When performing a linear regression, there are 2 types of uncertainty in the prediction. First is the prediction of the overall mean of the estimate (ie the center of the fit). The second is the ...

The correct way to solve the 2 coincident problem is to calculate the probability of 2 people not sharing the same birthday month. For this example the second person has a 11/12 chance of not sharing ...

In the Monty Hall problem there is prior knowledge (on the part of the judges) of which door the goats are behind. If the contestant on the first pick happens to pick the door with the goat. The ...

There is 2 uncertainties here. As you mentioned there is the uncertainty with the slope thus the spreading curve at ends, but there is also an uncertainty at the mean. Yes, the curve is thinnest at ...

Your professor comments concerning the conditional mean is when x meets a particular condition. In this case the intercept is the conditional mean of y when x=0. If x never takes the value of 0, then ...

To perform the simulation, here is a one line solution using the sample function: sample(0:4, N, replace = TRUE, prob = c(0.1, 0.2, 0.4, 0.2, 0.1)) #where: # 0:4 is the sequence of values (0 to 4 ...

The total variablity is equal to the square root of the sum of the squares of the individual variabilities. So this this case: 56.32^2 = (35.75^2 + 44.26^2) Due to rounding error in the ...

In the formula, case one, is comparing the 'mpg' variable subsetted with the 'am' variable. In this case a vector of 19 values versus a vector of 13 values. In the non-formula example, one is ...

What should be done in this case? This depends on your current running conditions. Is x1 currently running close the old limit of 20? Does the experiment which has been done initially make still ...

You are fitting a linear vx but plotting it on a log scale. Try taking the log in your lm model fit and then plot. pm = lm(vy ~ poly(I(log(vx)), degree=5, raw=TRUE), data = dat)

Let us define $\mu_1= 383$ and $\mu_2= 168$ with reasonable small standard deviations. The differences between the means are over 4 standard deviations apart. By eye one can judge, with a large ...

The chance of A being $>2\sigma$ from the mean is $2.28\%$. pnorm(12, 10, 1, lower.tail = FALSE) Thus the chance of A & B (2 independent events both occurring) is equal to $A*B$ or $0.052\%$ ...

Yes, R's output multiple regression can be tricky to understand at first. Think about this way when pop =1 (the first categorical value) you can drop all of the terms with pop2 and pop3 so you linear ...

The effect estimate is the difference in the result from when your factor changes from the high value to the low value. In your problem statement above the effect estimate for factor A is the average ...

I think you are looking for something like this: df <- structure(list(day = 1:6, growth = c(0.036787944, 0.018348802, 0.0121861, 0.009104847, 0.007257658, 0.006027639)), ...

This a poorly worded question, but you have all of the information you need to solve this problem. Just for a quick estimate, a difference of 28 in the means with a standard deviation of 20 has a z ...

It looks like both A and K are binary factors so, I don't believe logistic regression is the best statistical test here. I would recommend Chisq or Fisher's Exact test. With that said, from your small ...

The formula to calculate the approximate confidence limits for a binomial test is: $z_{alpha/2}*\sqrt{p*q/n}$ In your case for a fair coin p = q = 0.5 and using $z_{alpha/2}=1.96$ for a 95% confidence ...

The t.test is for comparing only 2 conditions A vs B. Your problem is A vs B vs C. So one option is to perform 2 separate t.test and compare Q1 to Q4 for both Food types. The other option is to ...

Yes, since a X and Y are provided and paired=TRUE option is used. From help: "If both x and y are given and paired is TRUE, a Wilcoxon signed rank test of the null that the distribution of of x -...

When you have a 2 factor category independent variable and a continuous dependent variable then look at performing a t-test. In this case you have a categorical independent variable and count data ...

MichiganWater mention this is a split plot design, Temperature is the plot and the recipe is the subplot. Using R to design the experiment: library(agricolae) library(tidyr) Temp <- c("T1&...

Without a sample of your data here is a simple example. The basic workflow is creating the dataset, fitting the data, making the prediction and then plotting: #create fake data x<- seq(0, 10, 0.2) ...

Another potential way of solving this problem is with a Poisson Distribution. In this case $\lambda$ is 1/1000. So the probability of of observing 0 events with a single trial is: ppois(0, 1/1000) #...

Each draw is an independent random event, the odds of picking a unique balls is dependent on the number of previous picks. First draw 8/8 or 100% of drawing a unique ball Second draw is 7 out of 8 ...

To answer your question and expand on my comment: Since you have a matched pairs of before and after, then testing the difference is a better test than comparing the raw values. Also since you ...

If you plot(res.hc) you can see the tree is highly unbalanced and thus when you specify a cut of two, one is making the cut at the first branching point. In this case the tree has a single branch on ...

Sampling without replacement follows a hypergeometric distribution. Assuming you have a 5% defect rate. The chances of drawing a passing part on the first pick is equal to 228/240 or 95%, now on ...