Recent content by d3t3rt

Yes, you need to divide by the standard error.

Statdad, thank you for fixing my statement about known standard errors and distributional forms for the sample slope and intercept.

To truly understand why you use a standard Normal distribution to calculate p-values relating to \bar{X}, you first must understand the concept of a sampling distribution. The sampling distribution of \bar{X} is the distribution of values taken by the sample mean in all possible samples of the...

For a one-sided test with an alternative hypothesis that \mu is greater than some hypothesized value, the critical value, x in this case, is 1.645.

DrDu is right. If you want to see which model best describes the data, use AIC or BIC. These will use the log-likelihood value, but they will also take into account the number of estimated parameters. You cannot compare raw log-likelihoods between models. This is the straight answer you...

The likelihood function is the same as the joint pdf or pmf of your data. The only difference is the order of conditioning. In the likelihood, you condition on the observed data. The joint pdf/pmf conditions on the parameter(s).

You cannot calculate the probability of observing a range of parameters from the likelihood without proper normalization. Your interpretation is fallacious. Frequentists believe that parameter values are fixed, unknown quantities. If you really want to calculate probabilities of parameters...

Here is a website outlining many of Excel's shortcomings: http://www.cs.uiowa.edu/~jcryer/JSMTalk2001.pdf I am very suspect of the algorithms that Excel uses to calculate statistics. Very simple statistical summaries have been calculated incorrectly by Excel (e.g., sample standard...

How many explanatory variables do you have? What are you trying to model? Are you restricting your intercept to a value of zero? From the clear linear trend in your residual plot, a response transformation will not help your situation. What is the R code that you are using?

Formally, the Central Limit Theorem states for any population with a finite variance \sigma and mean \mu, for sufficiently large simple random sample of size n, the sample mean will follow a Normal distribution with mean \mu and standard error \sigma/\sqrt{n}. Generally, if your sample size is...

The standard Normal curve is symmetric and the area underneath any density curve is 1 (by definition). Therefore, it would be redundant to give areas under the curve going from a calculated z-score to negative infinity and also calculate the integral between your z-score and positive infinity...

A good reference for bootstrapping is Efron & Tibshirani (1993) An Introduction to the Bootstrap. They discovered the bootstrap. That said, I wish to address the inappropriateness of using a bootstrap to find the standard error of the slope and intercept of a simple linear regression. The...