Final_Review_1_Solutions

5 calculate the p value and determine if the data

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: s a randomization test. He wrote the students’ grades on 50 index cards, shuffled the cards and dealt them into two groups: cheat sheet and open book. He then recorded the difference between the two group averages (¯cheat sheet − xopen book ) and x ¯ repeated 100 times. The dot plot below shows the distribution of these differences (denoted as diffMean). 5 Calculate the p-value and determine if the data provide evidence to support the professor’s theory. p-value = P(observed or more extreme outcome | H0 true) = P(xcheat sheet − xopen book ≥ 6 | µcheat sheet = µopen book ) ¯ ¯ = 3 / 100 = 0.03 Yes, with p-value < 0.05, we reject H0 . The data provide convincing evidence to support the professor’s theory that students who take the exam with just a cheat sheet score better on average. 9. We are interested in estimating the proportion of graduates at a mid-sized university who found a job within one year of completing their undergraduate degree. We conduct a survey and find out that out of the 400 randomly sampled graduates 348 found jobs within within one year of completing their undergraduate degree. (a) Define the sample statistic and the population parameter of interest. What is the value of the sample statistic? The sample statistic is the proportion of graduates in the sample who found a job within one year of graduating, p = 348 = 0.87. The population parameter of interest is the proportion of all ˆ 400 graduates from this university who found a job within one year of graduating, p. (b) Are assumptions and conditions for inference satisfied? 1. Independence Assumption: • Random Sampling Condition: We are told that the sample is random. • 10% Condition: We can safely assume that 400 < 10% of all students at a mid-sized university. Since we have a random sample and the 10% condition is satisfied, we can assume that whether or not one student in the sample has found a job is independent of another. 2. Nearly Normal Condition: First we must check if the success failure condition is met. np ≥ 10 → 400 ∗ 0.87 = 348 > 10￿ ˆ nq ≥ 10 → 400 ∗ 0.13 = 52 > 10￿ ˆ Since the observations are independent and the success-failure condition is met, we can assume that p is nearly normal. ˆ 6 (c) Construct a 95% confidence interval for the proportion of graduates who who found a job within one year of completing their undergraduate degree at this university. A 95% confidence interval can be calculated as follows: ￿ ￿ p(1 − p) ˆ ˆ 0.87 × 0.13 ∗ p±z ˆ = 0.87 ± 1.96 × n 400 = 0.87 ± 0.033 = (0.837, 0.903) (d) Explain what this interval means in the context of this question. We are 95% confident that the true proportion of graduates from this university who found a job within one year of completing their undergraduate degree is between 83.7% and 90.3%. (e) What does “95% confidence” mean? 95% of random samples of 400 would produce a confidence interval that includes the true proportion of students at this university who...
View Full Document

This document was uploaded on 12/04/2013.

Ask a homework question - tutors are online