Question

random digit dialing of residential numbers was done using a database of exchanges. a total of 2461 persons were contacted, of which 726 completed the survey. when a household was reached, surveyors asked to speak to the adult over 18 with the next birthday. no cell phone numbers were included in the provided data on their ages. 253 were over 65 years old.
according to u.s. census data, 13% of the population of adults in greenville county are over 65 years old. if the 689 completed surveys that included age data are an srs of adult greenville county residents, do you feel that the difference between the proportion of those over 65 that completed the survey and the population proportion of 13% can be easily explained by chance variation?
select the correct hypotheses, p - value, and conclusion in the context of the problem.
for this final exam, you will submit the work for this question within question 29. for this question, you will need to submit crunchit, or a ti calculator output from these technologies in jpeg/ jpg format, properly labeled, within your submission. only these technologies are allowed - no other excel spreadsheets should be used. upload the labeled output as part of your submission in a pdf or word document that contains your answers to all 28 questions. failure to include the required technology output may result in the loss of points.
\\(\circ\\) a \\(h_{0}: p_{1}=p_{2}\\) and \\(h_{a}: p_{1}\
eq p_{2}\\), where \\(p_{1}\\) is the proportion of seniors in all of greenville county and \\(p_{2}\\) is the proportion of seniors in the sample. \\(p = 0.0000\\). this is significant evidence that the sample does not represent the population of greenville in terms of age.
\\(\circ\\) b \\(h_{0}: p = 0.13\\) and \\(h_{a}: p\
eq 0.13\\) \\(p = 0.0000\\) there is significant evidence that the sample does not represent the population of greenville in terms of age.
\\(\circ\\) c \\(h_{0}: p = 0.13\\) and \\(h_{a}: p\
eq 0.13\\) \\(p = 1.0000\\) this is not enough evidence that the sample represents the population of greenville in terms of age.
\\(\circ\\) d \\(h_{0}: p = 0.13\\) and \\(h_{a}: p\
eq 0.13\\) \\(p = 0.0000\\) this is not enough evidence that the sample represents the population of greenville in terms of age.

Explanation:

Step1: Define Proportions

Let \( p_1 \) be the population proportion (13% or \( 0.13 \)) and \( p_2 \) be the sample proportion. The sample size \( n = 689 \), and the number of seniors in the sample \( x = 253 \). So, \( p_2=\frac{253}{689}\approx0.367 \).

Step2: Hypothesis Testing Setup

We test \( H_0: p_1 = p_2 \) (sample represents population) vs \( H_a: p_1
eq p_2 \) (sample does not represent population). Calculate the test statistic and p - value. The p - value here is given as \( 0.0000 \), which is less than typical significance levels (e.g., \( 0.05 \)).

Step3: Interpret Results

A very small p - value ( \( p = 0.0000 \)) indicates strong evidence against \( H_0 \). The sample proportion \( \hat{p}=\frac{253}{689}\approx0.367 \), and the population proportion \( p_1 = 0.13 \). The difference \( \hat{p}-p_1\approx0.367 - 0.13=0.237 \), but the option b has \( \hat{p}=0.13 \)? Wait, no, re - reading: Wait, the sample proportion of seniors is \( \frac{253}{689}\approx0.367 \), but the population proportion is \( 0.13 \). Wait, the options: Option b says \( \hat{p}
eq0.13 \) (which is true, since \( 0.367
eq0.13 \)) and \( P = 0.0000 \), indicating significant evidence that the sample does not represent the population. Option a says \( \hat{p}=p_2 \) (which is a mis - notation, \( p_2 \) should be sample proportion, \( p_1 \) population). Option c and d say not enough evidence, but p - value is 0.0000, so there is enough evidence. So option b is correct as it states there is significant evidence the sample does not represent the population (since \( \hat{p}
eq0.13 \) and p - value is very small).

Answer:

b. \( \hat{p}
eq0.13 \) and \( P = 0.0000 \) There is significant evidence that the sample does not represent the population of Greenville in terms of age.