- Find the sample proportion:
p= 9/75 = 0.12
Use 90% z-value: z^*=1.645
Standard error:
sqrt {0.12(0.88)/75} = 0.0375
Margin of error:
- 1.645(0.0375) = 0.062
Confidence interval:
- 0.12 +/- 0.062 = (0.058, 0.182)
Answer: (A) 0.058<p<0.182
Sandra B.
asked 04/20/21The president of an audio technology company is alarmed by the declining sales of the company's MP3 players. To better understand market trends, company employees asked 75 randomly chosen teenagers whether they currently own MP3 players; 9 answered YES, 66 answered NO. Find a 90% confidence interval for p, the proportion of teenagers who own MP3 players.
(A) 0.058<p<0.182
(B) 0.088<p<0.188
(C) 0.055<p<0.128
p= 9/75 = 0.12
Use 90% z-value: z^*=1.645
Standard error:
sqrt {0.12(0.88)/75} = 0.0375
Margin of error:
Confidence interval:
Answer: (A) 0.058<p<0.182
Get a free answer to a quick problem.
Most questions answered within 4 hours.
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.