Professor York randomly surveyed 240 students at Oxnard University, and found that 150 of the students surveyed watch more than 10 hours of television weekly. Develop a 95% confidence interval to estimate the true proportion of students who watch more than 10 hours of television each week. The confidence interval is:
A. .533 to .717
B. .564 to .686
C. .552 to .698
D. .551 to .739
Use confidence interval formulas for proportions.
CI95 = p + or - (1.96)(√pq/n)
...where √ = square root, p = x/n, q = 1 - p, and n = sample size.
Hint: x = 150, n = 240
Convert all fractions to decimals to work the formulas.
I hope this will help get you started.
^^^ not a good answer
The answer is B
0.625 +/- (1.960) times the square root of 0.0625(1-0.0625)/240
To develop a confidence interval to estimate the true proportion of students who watch more than 10 hours of television each week, we can use the following formula:
CI = p̂ ± z * sqrt((p̂ * (1 - p̂)) / n)
Where:
- p̂ is the sample proportion (150/240 = 0.625)
- z is the z-score for the desired confidence level (95% confidence level corresponds to a z-score of approximately 1.96)
- n is the sample size (240)
Let's calculate the confidence interval:
CI = 0.625 ± 1.96 * sqrt((0.625 * (1 - 0.625)) / 240)
= 0.625 ± 1.96 * sqrt(0.234375 / 240)
= 0.625 ± 1.96 * sqrt(0.000976563)
Now, let's calculate the upper and lower bounds of the confidence interval:
Lower bound = 0.625 - 1.96 * sqrt(0.000976563)
≈ 0.625 - 1.96 * 0.03125
≈ 0.625 - 0.0610
≈ 0.564
Upper bound = 0.625 + 1.96 * sqrt(0.000976563)
≈ 0.625 + 1.96 * 0.03125
≈ 0.625 + 0.0610
≈ 0.686
Therefore, the 95% confidence interval to estimate the true proportion of students who watch more than 10 hours of television each week is approximately 0.564 to 0.686.
So, the correct answer is B. .564 to .686.