Chapter 8 HW

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You must use software to solve these problems. Copy/paste the output to include with your answer. Answers without these results will receive zero credit. On the other hand, copy/pasted output without any interpretation will receive zero credit. Ch 8 homework problems (5 points each) 8.54 a – d 8.54 Where do you get your news? A report produced by the Pew Research Center’s Project for Excellence in Journalism summarized the results of a survey on how people get their news. Of the 2342 people in the survey who own a desktop or laptop, 1639 reported that they get their news from the desktop or laptop.26 a. Identify the sample size and the count. a. N=2342 b. Success count= 1639 c. Failure count= 703 b. Find the sample proportion and its standard error. a. P-hat= x/n= 1639/2342= .6998292 b. SE of p-hat= sqrt((p*(1-p))/n)= .009470803 c. Find and interpret the 95% confidence interval for the population proportion. a. [.6809519, .7180521] b. There is 95% confidence that the population proportion will fall within this interval. d. Are the guidelines for use of the large-sample confidence interval satisfied? Explain your answer. a. Yes because n is greater than 20, success count > 10, and failure count > 10 b. Also p-hat is not an extreme value because both the success and failure count is significantly above 10 reducing extremes. 1639/2342 and 703/2342 c. For a stable confidence interval these requirements must be satisfied and they all are 8.60 a – c 8.60 Facebook users. A Pew survey of 1802 Internet users found that 67% used Facebook.29 a. How many of those surveyed used Facebook? a. N=1802 b. # used Facebook= (1802)(.67)= 1207.34= 1207 b. Give a 95% confidence interval for the proportion of Internet users who used Facebook. a. [.6479482, .6913285] b. There is 95% confidence that the population proportion falls within this interval c. Convert the confidence interval that you found in part (b) to a confidence interval for the percent of Internet users who used Facebook. a. Since the proportions in the interval are between 0 and 1 multiplying the interval by 100 will give percentages

b. [64.8%, 69.1%] 8.74 a – c 8.74 Statistics and the law. Castaneda v. Partida is an important court case in which statistical methods were used as part of a legal argument.33 When reviewing this case, the Supreme Court used the phrase “two or three standard deviations” as a criterion for statistical significance. This Supreme Court review has served as the basis for many subsequent applications of statistical methods in legal settings. (The 2 or 3 standard deviations referred to by the Court are values of the z statistic and correspond to P-values of approximately 0.05 and 0.0026.) In Castaneda, the plaintiffs alleged that the method for selecting juries in a county in Texas was biased against Mexican Americans. For the period of time at issue, there were 181,535 persons eligible for jury duty, of whom 143,611 were Mexican Americans. Of the 870 people selected for jury duty, 339 were Mexican Americans. a. What proportion of eligible jurors were Mexican Americans? Let this value be p0. a. Po= 143611/181535= .7910926 b. Let p be the probability that a randomly selected juror is a Mexican American. The null hypothesis to be tested is H0: p=p0. Find the value of p^ for this problem, compute the z statistic, and find the P-value. What do you conclude? (A finding of statistical significance in this circumstance does not constitute proof of discrimination. It can be used, however, to establish a prima facie case. The burden of proof then shifts to the defense.) a. P-hat= 339/870= .3896552 b. X-squared= 848.35 -> test stat= sqrt(848.35)= 29.12645 c. P-value< 2.2e-16 d. P-value is approximately 0 meaning < .05 ; reject the null/ Ho the proportions show significant difference i. The Po= 79% and p-hat= 38.97% shows a wide gap further expressing the small p-value which are nowhere near equal c. We can reformulate this exercise as a two-sample problem. Here we wish to compare the proportion of Mexican Americans among those selected as jurors with the proportion of Mexican Americans among those not selected as jurors. Let p1 be the probability that a randomly selected juror is a Mexican American and let p2 be the probability that a randomly selected nonjuror is a Mexican American. Find the z statistic and its P-value. How do your answers compare with your results in part (b)? a. N1= 870 and x1=339 b. N2= 181535-870= 180665 and x2= 143611-339= 143272 c. X-squared= 852.43 z-test stat= sqrt(852.43)= 29.1964 d. P-value < 2.2e-16 e. These results compare to part b because they have the same p-value and pretty much the same z-test stat. Both tests reject Ho as a result, there is significant differences within proportions.

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