Question 1189975: 1.We wish to estimate what percent of 18 to 25-year old's that own a Samsung phone. Out of 400 sampled, 104 had a Samsung phone. Based on this, construct a 90% confidence interval for the proportion of all adults that own a Samsung phone.
2. A 2011 Gallup survey based on telephone and face-to-face interviews with a sample of adults in China suggests that 20% smoke regularly or occasionally, with a margin of error of 2.2%. Give the confidence interval.
3. You want to obtain a sample to estimate the average number of hours per week community college students in the United States work at their job while being a full-time student. Based on previous evidence, you believe the population standard deviation is 10. You would like to be 99% confident that your estimate is within 1.5 hours of the true population mean.
Assume that the number of statistics students is unknown, but very large. How large of a sample size is required?
4.You want to obtain a sample to estimate the average number of hours per week Nexford students work at their job while being a full-time student. Based on previous evidence, you believe the population standard deviation is 10. You would like to be 99% confident that your estimate is within 1.5 hours of the true population mean.
Assume that the number of students at Nexford is 2900 students. How large of a sample size
is required?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
In the future, please only post one problem at a time.
I'll do problem 1 to get you started.
n = 400 = sample size
x = 104 = number who have a Samsung phone
phat = x/n = 104/400 = 0.26
The sample proportion phat estimates the population proportion p.
At 90% confidence, the z critical value is roughly z = 1.645 (use a table or calculator to find this value).
We can use the Z distribution because n > 30.
Let's calculate the lower and upper bounds of the confidence interval.
L = lower bound of confidence interval
L = phat - z*sqrt(phat*(1-phat)/n)
L = 0.26 - 1.645*sqrt(0.26*(1-0.26)/400)
L = 0.26 - 0.03607766656811
L = 0.26 - 0.036
L = 0.224
U = upper bound of confidence interval
U = phat + z*sqrt(phat*(1-phat)/n)
U = 0.26 + 1.645*sqrt(0.26*(1-0.26)/400)
U = 0.26 + 0.03607766656811
U = 0.26 + 0.036
U = 0.296
The confidence interval is in the format of (L, U)
Answer: (0.224, 0.296)
We're 90% confident that the population proportion p is somewhere between 0.224 and 0.296
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