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of+Group Take a Test- Seong Ho Kang- Google Chrome a Secure I https:/www.mathxl.com/Student/PlayerTest.aspx?testid-1846180168centerwinsyes O+blood+ o- About 329,000 res Math 165 Summer 2018 t (and any subsequ Seong Ho Kang &I; 7/22/18 9:22 AM Assume 42% Test: Test 3 Chapters 6 and7 https://www.cheal A hospital is condu blood is low, and it This Question: 1 pt estimate the proba Time Remaining: 01 3054 Submit Tes 35 of 60 (33 complete) This Test: 60 pts possible Solved: Assu In a survey of 1248 people, 797 people said they voted in a recent p esden al election Votrg ecords show that 61% of elgble voters actualy dd de G ent 61% ofe gble des actual did ntps/www cheqvot() the probablity that among 1248 randomly selected voters, at least 797 actual did vote (b) What do the results from part (a) suggest? A hospital is its supply of Group probability that the (a) PI2797) (Round to four decimal places as needed) (b) What does the result from part (a) suggest? OA. People are being honest because the probabity of P(x2797)is less than 5% O B. People are being honest because the probability ofPO 2797)is atleast 1% Probabilities https/answers.y B Mar 18, 2013 it's supply of g .-.]| C. D. Some people are being less than honest because P(x 2 797) is less than 5% Some people are being less than honest because P(x 2797) is at least 1 % Habits, hassle https://www.rese 7.4 7 TO WHAT EX Donor deferral due thus reliant on a the US Porl Statistics https://www.note Jul 13,20TO mat not be concerned has some strilking some exam Por military bl militaryblood dod fici blood, especiallyExplanation / Answer
The sample size of the survey, n is 1248.
Out of the 1248 people, 797 said they voted in the elections. This is the sample proportion.
Thus, sample proportion, p = 797/1248= 0.6386218= 0.64
The population proportion, P= 0.61
Given that this information is true, we need to find the probability that 797 people did actually vote. (In the sample of 1248)
Let us first find the Z-score of this observation.
The standard error of this observation is sqrt((p*q)/n)= sqrt(0.64*0.36/ 1248)
= sqrt(0.6386218)= 0.01358732
Thus, the z-score is (p-P)/ SE= (0.64-0.61)/0.01358732= 2.207941
The probability of X>=797= 1- probability of X<797= 1- 0.9864
=0.0136= 1.36%
The second question, in my opinion, can have more than 1 answers.
Since the probability of X being greater than 797 is very small, we may say that people are being honest about voting, because not many people above 797 seem to have voted (probability being less than 5%).
On the other hand, since the probabilty of X being greater than 797 is at least 1%, we may say that some people are being less than honest (since the required probability is at least 1%).
I believe the first option seems to be more right. 5% is a good p-value. 1% is too less a p-value to judge the honesty of people.
Thus, 1.36% suggests that people are being honest about voting, because not many people above 797 seem to have voted (probability being less than 5%).
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