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Chapter 5: Problem 38

Randomization Slopes A randomization distribution is created to test a null hypothesis that the slope of a regression line is zero. The randomization distribution of sample slopes follows a normal distribution, centered at zero, with a standard deviation of 2.5 (a) Draw a rough sketch of this randomization distribution, including a scale for the horizontal axis. (b) Under this normal distribution, how likely are we to see a sample slope above \(3.0 ?\) (c) Find the location of the \(5 \%\) -tile in this normal distribution of sample slopes.

Short Answer

Expert verified
a) The sketch should show a bell curve centered at zero with standard deviations marked as 2.5, 5.0, -2.5, -5.0, etc. b) The exact probability depends on the values in the standard normal distribution table, but it is small. c) The 5% percentile of this distribution, in terms of the slope of the regression line, again depends on the exact values in the table but is a negative value.

Step by step solution

01

Sketch the Randomization Distribution

The distribution is a normal distribution with mean 0 and standard deviation 2.5. On a set of axes, sketch a bell-curve symmetrical about the vertical axis. This represents a normal distribution. Mark the mean (0) in the center of the horizontal axis and indicate the standard deviations on either side of it. Indicate positive values on the right and negative values on the left of the mean.
02

Calculating the Probability of a Sample Slope Above 3.0

To find the probability that the sample slope is above 3.0, first calculate the Z-score of 3.0 with a mean of 0 and standard deviation of 2.5 using the formula: Z= \( \frac{x-\mu}{\sigma} \). Look up the Z score in a standard normal distribution table to find the probability. Since we want the probability of the sample slope being above 3.0, subtract the table value from 1.
03

Finding the 5% percentile of the Distribution

Find the Z-score that corresponds to the 5% percentile in a standard normal distribution table. Transform this Z-score back to our normal distribution with mean 0 and standard deviation 2.5 using the formula: X = Z * \( \sigma \) + \( \mu \) . This gives the 5% percentile in terms of the slope of the regression line.

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Most popular questions from this chapter

Curving Grades on an Exam A statistics instructor designed an exam so that the grades would be roughly normally distributed with mean \(\mu=75\) and standard deviation \(\sigma=10 .\) Unfortunately, a fire alarm with ten minutes to go in the exam made it difficult for some students to finish. When the instructor graded the exams, he found they were roughly normally distributed, but the mean grade was 62 and the standard deviation was 18\. To be fair, he decides to "curve" the scores to match the desired \(N(75,10)\) distribution. To do this, he standardizes the actual scores to \(z\) -scores using the \(N(62,18)\) distribution and then "unstandardizes" those \(z\) -scores to shift to \(N(75,10)\). What is the new grade assigned for a student whose original score was \(47 ?\) How about a student who originally scores a \(90 ?\)

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