{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

chapter 6 homework manager

chapter 6 homework manager - Question 1 Score 1/1 Assuming...

This preview shows pages 1–5. Sign up to view the full content.

Question 1: Score 1/1 Assuming the same level of significance , as the sample size increases, the value of t /2 approaches the value of z /2 . Your Answer: Choice Selecte d True False Question 2: Score 0/1 The width of a confidence interval will be: Your Answer:

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Choice Selected Correct Narrower for 99% confidence than 95% confidence. Wider when the sample standard deviation (s) is small than when s is large. Narrower for 90% confidence than 95% confidence. Wider for a sample size of 100 than for a sample size of 50. Question 3: Score 1/1 If we have a sample size of 100 and the estimate of the population proportion is .10, we can estimate the sampling distribution of with a normal distribution. Your Answer:
Choice Selecte d True Suppose that we will randomly select a sample of n = 90 units from a population and that we will compute the sample proportion of these units that fall into a category of interest. If the true population proportion p equals 0.9 : a. Describe the shape of the sampling distribution of . Why can we validly describe the shape? b. Find the mean and the standard deviation of the sampling distribution of . c. Calculate the following probabilities about the sample proportion . In each case sketch the sampling distribution and the probability. 1. P ( > .96) 2. P (.855 < < .945) 3. P ( < .915) Round answer to part b to 4 decimal places. Round z-values to 2 decimal places. a. np = ( 90 )(.9) = 81 : n(1 - p) = ( 90 )(.1) = 9 : The sample is large enough The distribution of is normally distributed. b. σ = 0.0316 c. (1) P ( > .96) = 0.0287 with a tolerance of ±0.004 (2) P (.855 < < .945) = 0.8444 with a tolerance of ±0.004 (3) P ( < .915) = 0.6808 with a tolerance of ±0.004 On January 4, 2000, the Gallup Organization released the results of a poll concerning public skepticism about the extent of Y2K computer problems. The poll results were based on a randomly selected national sample of 642 adults, 18 years and older, conducted December 28, 1999. One question asked if the respondent felt that media warnings about possible Y2K computer problems were "necessary precautions." a. Suppose that we want to justify the claim that a majority of U.S. adults believe that media warnings about possible Y2K computer problems were necessary precautions. The poll actually found that 65 percent of the respondents felt this way. If, for the sake of argument, we assume that 54 percent of U.S. adults believe that the media warnings were necessary precautions (that is, p = 0.54), calculate the probability of observing a sample proportion of 0.65 or more. That is, calculate P ( > 0.65). b. Based on the probability you computed in part a, would you conclude that a majority of U.S. adults believe that the media warnings were necessary precautions? Explain.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Round your answers to 2 decimal places. a. µ = 0.54 σ = 0.02 P ( > 0.65) is less than .001 b. Yes , the probability of observing this sample is extremely small if p = 0.54 . More likely the true p is greater than 0.54 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}