If the alpha level changed from a=0.05 to a=0.01, what happens to the critical region and ehat happens to the probability of type 1 error

A random sample of 28 records of automobile driver fatalities in kit carson coybry showed 15 involved an intoxicated driver. Do these data indicate that the population proportion of driver fatalities related to alochol is less than 77%. Use a=0.01 to determine the test statistic, the critical value, and p value.

the standard normal distribution is just a normal distribution with μ = 0 and σ^2 = 1. it turns out that if we standardize a normal random variable X with parameters μ, σ in the following way   z = x-μ/σ the expected value of the transformation is 0 and variance is 1. so if Z = 0.2, when μ=3 and σ= 0.2, what is the normal random variable X has value?

You are studying a gene that controls horn lergth in cows. The wild-type long-horn allele (L) is dominant to the mutant short-horn (l) allele. However, tne L allele is only 60 penetrant, You cross two heterozygous cows. What proportion of offspring would you expect to exhibit the long-horn nhenoype? Assume the penetrance of Lapplies equelly te both homozygotes and heterozygotes and that there is no difference observed between 7. the sexes. CA. 0.40 O8.0.75 O C.D55 O D.0.45 OE.0.60 Reset Selecion Mark for Rrview What's Thist

Suppose you run a regression y=alpha + beta*x + u. You know that the model sum of squares equal 2998.9, residual sum of squares equals 571,612.8, and residual mean squares (MS) squals 412.41. What is the number of observations in your sample?

Heteroskedasticity arises because of non-constant variance of the error terms. We said proportional heteroskedasticity exists when the error variance takes the following structure: Var(et)=σt^2=σ^2 xt. But as we know, that is only one of many forms of heteroskedasticity. To get rid of that specific form of heteroskedasticity using Generalized Least Squares, we employed a specific correction – we divided by the square root of our independent variable x. And the reason why that specific correction worked, and yielded a variance of our GLS estimates that was sigma-squared, was because of the following math: (Picture 1) Where var(et)=σ^2  according to our LS assumptions. In other words, dividing everything by the square root of x made this correction work to give us sigma squared at the end of the expression. But if we have a different form of heteroskedasticity (i.e. a difference variance structure), we have to do a different correction to get rid of it. (a) what correction would you use…

1) If there is a 5% chance of a Type I error what is the alpha value (α)? 2) If there is a 20% chance of a Type II error what is the beta value (β)?