MATH 225N Week 6 Confidence Intervals Questions and Answers

1. On a busy Sunday morning, a waitress randomly sampled customers about their preference for morning
beverages, Specifically, she wanted to find out how many people preferred coffee over tea. The proportion
of customers that preferred coffee was 0.42 with a margin of error 0.07.
Construct a confidence interval for the proportion of customers that preferred coffee.
(0.42 - 0.07), (0.42 + 0.07) = (0.35), (0.49)
2. A company sells juice in 1quart bottles. In a quality control test, the company found the mean volume of
juice in a random sample of bottles was X = 31 ounces, with a marginal error of 3 ounces.
Construct a confidence interval for the mean number of ounces of juice bottled by this company.
(31-3), 31+3) = (28), (34)
3. Randomly selected employees at an office were asked to take part in a survey about overtime. The office
manager wanted to find out how many employees worked overtime in the last week. The proportion of employees
that worked overtime was 0.83, with a margin of error of 0.11.
(0.83 – 0.11), (0.83 + 0.11) = (0.72), (0.94)
4. A random sample or garter snakes were measured, and the proportion of snakes that were longer than 20
inches in length recorded. The measurements resulted in a sample proportion of p = 0.25 with a sampling
standard deviation of Op = 0.05.
Write a 68% confidence interval for the true proportion of garter snakes that were over 20 inches in length.
(.25 - .05). (.25 + .05) = (0.20), (.30)
5. The average number of onions needed to make French onion soup from the population of recipes is unknown.
A random sample of recipes yields a sample mean of x = 8.2 onions. Assume the sampling distribution of the
mean has a standard deviation of 2.3 onions.
Use the Empirical Rule to construct a 95% confidence interval for the true population mean number of onions.
Since 95% falls in 2 SD’s the calculation would be (8.2 – 4.6) “4.6 is the margin of error”, (8.2 + 4.6) = (3.6) , (12.8)
6. In a survey, a random sample of adults were asked whether a tomato is a fruit or vegetable. The survey resulted
in a sample proportion of 0.58 with a sampling standard deviation of 0.08 who stated a tomato is a fruit.
Write a 99.7 confidence interval for the true proportion of number of adults who stated the tomato is a fruit.
(0.58 – 3 x 0.08), (0.58 + 3 x 0.08) =(0.58 – .24), (0.58 + .24) = 0.34 + 0.82
7. A college admissions director wishes to estimate the mean number of students currently enrolled. The age of
random sample of 23 students is given below. Assume the ages are approximately normally distributed. Use Excel
to construct a 90% confidence interval for the population mean age. Round your answer to 2 decimal places and
use increasing order.
Use week 6 worksheet to get mean and SD.
Data
25.8 Mean 23.1043
22.2 Sample Standard Deviation 1.3693